Background material from Algebra II

0.1 Rings and polynomials

Exercise 0.1 $\;$Find the multiplicity of the root $x_0=2$ of the polynomial \[f(x)=x^5-5x^4+7x^3-2x^2+4x-8\in\mathbb{Q}[x].\]

Solution

We could just repeatedly factor out $(x-2)$ terms as many times as we can: \[\begin{align*} x^5-5x^4+7x^3-2x^2+4x-8 &= (x-2)(x^4-3x^3+x^2+4) \\ &= (x-2)^2(x^3-x^2-x-2) \\ &= (x-2)^3(x^2+x+1) \end{align*}\] and since $x_0=2$ is not a root of $x^2+x+1$, we find the multiplicity is $3$.

Alternatively, we could use the derivative criterion: a root $x_0$ has multiplicity $r$ if and only if \[f(x_0)=f'(x_0)=\cdots=f^{(r-1)}(x_0)=0\quad\text{and}\quad f^{(r)}(x_0)\neq 0.\] In this case we have

$f(2)=0$ so $x_0=2$ is a root (i.e. the multiplicity is at least $1$).
$f'(x)=5x^4-20x^3+21x^2-4x+4$ and $f'(2)=0$ so the multiplicity is at least $2$.
$f''(x)=20x^3-60x^2+42x-4$ and $f''(2)=0$ so the multiplicity is at least $3$.
$f'''(x)=60x^2-120x+42$ and $f'''(2)=42\neq 0$ so the multiplicity is exactly $3$.

Exercise 0.2 $\;$Find a value of $a\in\mathbb{Q}$ such that $x_0=-1$ is a root of \[f(x)=x^5-ax^2-ax+1\in\mathbb{Q}[x].\] of multiplicity at least $2$.

Solution

We just need \[f(-1)=-1-a+a+1=0\quad\text{and}\quad f'(-1)=5+2a-a=0.\] This happens only when $a=-5$.

We can easily check that $x^5+5x^2+5x+1=(x+1)^2(x^3-2x^2+3x+1)$. Also $-1$ is not a root of the cubic term, so it has multiplicity exactly $2$.

Exercise 0.3 $\;$Consider the degree $n$ truncated exponential polynomial \[f_n(x)=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}\in\mathbb{Q}[x].\] For any given $n\geq 1$, show that $f_n(x)$ does not have any multiple roots.

Solution

If $x_0$ is a root of multiplicity at least $2$, then $f_n(x_0)=f_n'(x_0)=0$, i.e. \[\begin{align*} f_n(x_0)&=1+\frac{x_0}{1!}+\frac{x_0^2}{2!}+\cdots+\frac{x_0^n}{n!} = 0 \\ \text{and}\quad f_n'(x_0)&=1+\frac{x_0}{1!}+\frac{x_0^2}{2!}+\cdots+\frac{x_0^{n-1}}{(n-1)!} =0 \quad \end{align*}\] Subtracting the second equation from the first gives $x_0^n/n!=0$ and so $x_0=0$.

However, $f_n(0)=1\neq 0$ so there are no solutions to both equations and $f_n(x)$ has no multiple roots.

Exercise 0.4 $\;$Find the greatest common divisor $\gcd(f(x),g(x))$ for the following pairs of polynomials $f(x), g(x)\in\mathbb{F}_2[x]$ over the field of $2$ elements :

$\;(a)\;$ $f(x)=x^5+x^4+1\;$ and $\;g(x)=x^4+x^2+1$.

$\;(b)\;$ $f(x)=x^5+x^3+x+1\;$ and $\;g(x)=x^4+1$.

Solution

$(a)\;$ Apply Euclid’s algorithm for polynomials (remembering that $2=0$ in the field $\mathbb{F}_2$): \[\begin{align*} x^5+x^4+1 &= (x+1)(x^4+x^2+1)+(x^3+x^2+x) \\ x^4+x^2+1 &= (x+1)(x^3+x^2+x)+(x^2+x+1) \\ x^3+x^2+x &= x(x^2+x+1) \end{align*}\] Hence we find $\gcd(f,g)=x^2+x+1$.

$(b)\;$ Similarly \[\begin{align*} x^5+x^3+x+1 &= x(x^4+1)+(x^3+1) \\ x^4+1 &= x(x^3+1)+x+1 \\ x^3+1 &= (x^2+x+1)(x+1) \end{align*}\] Hence this time we find $\gcd(f,g)=x+1$.

Exercise 0.5 $\;$Which of the following sets of complex numbers form a ring under the usual addition and multiplication of $\mathbb{C}$? Give reasons.

$\;(a)\;\;\mathbb{Z},\hspace{8em}$ $\;(b)\;\;\mathbb{Z}_{\geq 0},\hspace{8em}$ $\;(c)\;\;\mathbb{Q},$

$\;(d)\;\;\left\{a+b\,\sqrt{2} \;|\; a, b\in\mathbb{Q}\right\},\hspace{4.1em}$ $\;(e)\;\;\left\{a+b\,\sqrt[3]{2} \;|\; a, b\in\mathbb{Q}\right\},$

$\;(f)\;\;\left\{a+b\,i\;|\;a,b\in\mathbb{Q}\right\},\hspace{5.3em}$ $\;(g)\;\;\left\{a+b\,i\;|\;a,b\in\mathbb{Z}\right\},$

$\;(h)\;\;\left\{a+b\,\sqrt[3]{2}+c\,\sqrt[3]{4} \;|\; a, b, c\in\mathbb{Q}\right\},$

$\;(i)\;\;\left\{\frac{a}{p^n}\;|\;a\in\mathbb{Z}, n\in\mathbb{Z}_{\geq 0}\right\}$ where $p$ is a fixed prime number.

Solution

Since these are all subsets of the ring $\mathbb{C}$ and have the same operations, all we need to do is check that they contain $0$ and are closed under addition and multiplication

$(a)\;$ Yes. (It’s the prototypical example of a ring!)

$(b)\;$ No. The set doesn’t have additive inverses e.g. $-1$ so even the addition breaks down.

$(c)\;$ Yes.

$(d)\;$ Yes.

$(e)\;$ No. It’s closed under addition but isn’t closed under multiplication. For instance, $\sqrt[3]{2}$ is in the set but $\left(\sqrt[3]{2}\right)^2=\sqrt[3]{4}$ isn’t. This is true but perhaps it isn’t obvious? See if you can prove that there are no rational numbers $a,b\in\mathbb{Q}$ such that $\sqrt[3]{4}=a+b\sqrt[3]{2}$.

$(f)\;$ Yes.

$(g)\;$ Yes. These are the Gaussian integers.

$(h)\;$ Yes. We don’t have the same problem as in part $(e)$ since $\left(\sqrt[3]{2}\right)^2$, $\left(\sqrt[4]{2}\right)^2=2$ and $\sqrt[3]{2}\sqrt[3]{4}$ are all in the set.

$(i)\;$ Yes. Just notice sums and products of the form $a/p^n$ are of the same form.

Exercise 0.6 $\;$Which of the following subsets of functions $f:\mathbb{R}\rightarrow\mathbb{R}$ form a ring under the usual addition and multiplication of functions? Give reasons.

$\;(a)\;\;$ All continuous functions on a fixed closed interval $[a,b]$,

$\;(b)\;\;$ All twice differentiable functions on a fixed open interval $(a,b)$,

$\;(c)\;\;$ All rational-valued functions $f:\mathbb{R}\rightarrow\mathbb{Q}$,

$\;(d)\;\;$ All functions which equal $0$ on a given subset $D\subset\mathbb{R}$.

Solution

The set of all functions $f:\mathbb{R}\to\mathbb{R}$ is a ring, so we can use the subring test: just check the subset contains the zero function and is closed under subtraction and multiplication. Strictly speaking, since we expect rings to include a multiplicative identity, we should check for this aswell.

$(a)$ If $f(x), g(x)$ are continuous functions on $[a,b]$ then so are $f(x)\pm g(x)$ and $f(x)g(x)$. So are the additive and multiplicative identity functions $f(x)\equiv 0$ and $f(x)\equiv 1$.

$(b)$ If $f(x), g(x)$ are twice-differentiable functions on $(a,b)$ then so are $f(x)\pm g(x)$ and $f(x)g(x)$. So are $f(x)\equiv 0$ and $f(x)\equiv 1$.

$(c)$ If $f(x), g(x)$ are rational-valued functions then so are $f(x)\pm g(x)$ and $f(x)g(x)$. So are $f(x)\equiv 0$ and $f(x)\equiv 1$.

$(d)$ If $f(x), g(x)$ equal $0$ on $D$ then so do $f(x)\pm g(x)$ and $f(x)g(x)$. So does $f(x)\equiv 0$ and this is the additive identity. But $f(x)\equiv 1$ doesn’t equal $0$ on $D$ so how can we have a multiplicative identity? We can define it by $f(x)=0$ on $D$ and $1$ on $\mathbb{R}\setminus D$.

Exercise 0.7 Determine all zero divisors in the direct product $\mathbb{C}\times\mathbb{C}$, i.e. the ring of all pairs $(a,b)$ with $a, b\in\mathbb{C}$ and addition and multiplication given by \[(a,b)+(c,d)=(a+b,c+d)\quad\text{and}\quad (a,b)\cdot(c,d)=(ac,bd).\]

Solution

The zero element in $\mathbb{C}\times\mathbb{C}$ is $(0,0)$. Suppose that $(a,b)\neq(0,0)$.

If $a\neq 0$ and $b\neq 0$, then \[(a,b)\cdot(c,d)=(ac,bd)=(0,0)\quad\implies\quad (c,d)=(0,0)\] and so $(a,b)$ is not a zero divisor.
If $a\neq 0$ then $(a,0)\cdot(0,1)=(0,0)$ so $(a,0)$ is a zero divisor.
If $b\neq 0$ then $(0,b)\cdot(1,0)=(0,0)$ so $(0,b)$ is a zero divisor.

Exercise 0.8 $\;(a)\;$ Find all group homomorphisms $\varphi:\mathbb{Z}\rightarrow\mathbb{Q}$ (with the usual additions).

$\;(b)\;$ Find all ring homomorphisms $\varphi:\mathbb{Z}\rightarrow\mathbb{Q}$ (with the usual additions and multiplications).

Solution

$(a)\;$ Choose any rational number $a\in\mathbb{Q}$. Then \[\begin{align*} \varphi\;:\; \mathbb{Z} &\longrightarrow \mathbb{Q} \\ n &\longmapsto na \end{align*}\] is a group homomorphism since $\varphi(m+n)=(m+n)a=ma+na=\varphi(m)+\varphi(n)$ for all $m, n\in\mathbb{Z}$.

Furthermore, any group homomorphism will be of this form since it will be uniquely determined by the image $a=\varphi(1)$ of the additive generator $1\in\mathbb{Z}$.

$(b)\;$ If $\varphi$ is a ring homomorphism, then it needs to satisfy $\varphi(m+n)=\varphi(m)+\varphi(n)$, so is again determined by $a=\varphi(1)$ as above.

However, we also require $\varphi(mn)=\varphi(m)\varphi(n)$ and this places a big restriction on the choices for $a$.

In particular, $a=\varphi(1)=\varphi(1)^2=a^2$ in $\mathbb{Q}$ and there are only two choices available:

$a=1$ gives $\varphi(n)=n$, i.e. the usual inclusion map of $\mathbb{Z}$ into $\mathbb{Q}$
$a=0$ gives $\varphi(n)=0$, i.e. the zero homomorphism mapping all elements to zero.

Note, only the first of these (the inclusion map) satisfies the extra condition $\varphi(1_{\mathbb{Z}})=1_{\mathbb{Q}}$ and so (under our definition) there is exactly one ring homomorphism $\mathbb{Z}\to\mathbb{Q}$.

A similar proof shows there is exactly one homomorphism $\varphi:\mathbb{Z}\to R$ for any ring $R$. It is given by \[\varphi(n)=\underbrace{1_R+...+1_R}_\text{$n$ times}\quad\text{for $n\geq 0$}\quad\text{and}\quad \varphi(n)=-\varphi(-n)\quad\text{for $n<0$.}\]

Exercise 0.9 $(\star)\;$ Recall in our definition of a ring homomorphism $\varphi:R\to S$, we require $\varphi(1_R)=1_S$.

Suppose we don’t assume this and only have $\varphi(a+b)=\varphi(a)+\varphi(b)$, $\varphi(a\cdot b)=\varphi(a)\cdot\varphi(b)$.

Show that if $S$ is an integral domain and $\varphi$ is not the zero homomorphism, then $\varphi(1_R)=1_S$ anyway.

Solution

We have \[ \varphi(1_R)\cdot\varphi(1_R)=\varphi(1_R\cdot 1_R)=\varphi(1_R)=1_S\cdot\varphi(1_R) \] \[\implies\quad \left[\varphi(1_R)-1_S\right]\cdot\varphi(1_R)=0_S.\] Now, if $S$ is an integral domain, this means that (at least) one of the terms being multiplied must be zero and so $\varphi(1_R)$ equals either $1_S$ or $0_S$. However, $\varphi(1_R)=0_S$ implies that for any $r\in R$ we have \[\varphi(r)=\varphi(1_R\cdot r)=\varphi(1_R)\cdot\varphi(r)=0_S\cdot\varphi(r)=0_S.\] In other words, $\varphi$ is the zero homomorphism. The only option left is $\varphi(1_R)=1_S$ as required.

Notice how we have used both of the extra conditions in proving this…

Exercise 0.10 $\;$Are the following pairs of quotient rings isomorphic? Give reasons.

$\;(a)\;$ $R_1=\mathbb{Z}[x]/(x^3+1)$ and $R_2=\mathbb{Z}[x]/(x^3+2x^2+x+1)$,

$\;(b)\;$ $R_1=\mathbb{Z}[x]/(x^2-2)$ and $R_2=\mathbb{Z}[x]/(x^2-3)$.

Solution

$(a)\;$ Notice that $x^3+1=(x+1)(x^2-x+1)$ is reducible. This means that the residues (i.e. the images in the quotient) of the factors $x+1$ and $x^2-x+1$ will be non-trivial zero divisors in $R_1$.

However, $R_2$ has no zero divisors since $x^3+2x^2+x+1$ is irreducible. (You can see this by checking that it has no linear factors by the rational root test - neither of $\pm 1$ are roots.)

This means that $R_1$ and $R_2$ are not isomorphic.

$(b)\;$ Let $\alpha$ be the image of $x$ in $R_1$ and $\beta$ be the image of $x$ in $R_2$. Then we can identify the rings via \[R_1\cong\left\{a+b\alpha \;|\; a, b\in\mathbb{Z}\right\}\quad\text{and}\quad R_2\cong\left\{a+b\beta \;|\; a, b\in\mathbb{Z}\right\}\] where the operations use the rules $\alpha^2=2$ and $\beta^2=3$.

Suppose $\varphi:R_1\to R_2$ is a ring isomorphism. Note that we must have $\varphi(1)=1$ by definition (or using the previous exercise since $\varphi$ is not the zero homomorphism and $R_2$ is an integral domain). Furthermore, we then have \[\varphi(\alpha)^2=\varphi(\alpha^2)=\varphi(2)=2\varphi(1)=2.\] On the other hand, if we write $\varphi(\alpha)=a+b\beta$ for some $a, b\in\mathbb{Z}$, then \[\varphi(\alpha)^2=(a+b\beta)^2=a^2+2ab\beta+b^2\beta^2=a^2+3b^2+2ab\beta.\] Equating these gives \[\left(a^2+3b^2\right)+\left(2ab\right)\beta=2 \quad\implies\quad a^2+3b^2=2\;\;\;\text{and}\;\;\; 2ab=0.\] But there are clearly no $a, b\in\mathbb{Z}$ satisfying these two equations. Hence $R_1$ and $R_2$ are not isomorphic.

0.2 Group theory

Exercise 0.11 $\;$Which of the following are groups under the given operation?

$\;(a)\;$ $\left(A,+\right)$ where $A$ is any of $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$,

$\;(b)\;$ $\left(A,\times\right)$ where $A$ is any of $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$,

$\;(c)\;$ $\left(A\setminus \{0\},\times\right)$ where $A$ is any of $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$,

$\;(d)\;$ $\left(n\mathbb{Z},+\right)$ for a given $n\in\mathbb{N}$,

$\;(e)\;$ $\left(\{-1,1\},\times\right)$,

$\;(f)\;$ $\left(\{a^n \;|\; n\in\mathbb{Z}\},\times\right)$ for fixed $a\in\mathbb{R}$,

$\;(g)\;$ $\left(\{z\in\mathbb{C} \;|\; z^n=1\},\times\right)$ for fixed $n\in\mathbb{N}$,

$\;(h)\;$ $\left(\{z\in\mathbb{C} \;|\; z^n=1\text{ for some } n\in\mathbb{N}\},\times\right)$,

$\;(i)\;$ $\left(\{z\in\mathbb{C} \;|\; |z|=R\} ,\times\right)$ for fixed $R>0$,

$\;(j)\;$ $\left(\{z\in\mathbb{C} \;|\; 0<|z|<R\} ,\times\right)$ for fixed $R>0$.

Solution

These are all subsets of complex numbers (with addition or multiplication) so associativity is automatic. We just need to check for closure, identity and inverses which in most cases are immediately clear…

$(a)\;$ These are all (very familiar!) groups except for $\mathbb{N}$ which doesn’t contain the additive identity $0$ or additive inverses (negative integers).

$(b)\;$ None of these are groups as there aren’t multiplicative inverses. For $\mathbb{N}$, we have e.g. $2^{-1}\not\in\mathbb{N}$. For the others, $0$ has no multiplicative inverse.

$(c)\;$ Again, $\mathbb{N}$ doesn’t have multiplicative inverses so not a group. All the others are groups (since it was only $0$ that was non-invertible.)

$(d)\;$ Yes. Closed since $na+nb=n(a+b)$, identity $0$ and inverses $-(na)=n(-a)$.

$(e)\;$ Yes, straightforward. You should recognise it is isomorphic to the cyclic group $C_2=\mathbb{Z}/2$.

$(f)\;$ Yes. Closed since $a^m\times a^n=a^{m+n}$, identity is $a^0=1$ and inverses $(a^n)^{-1}=a^{-n}$.

Actually when $a=-1$ this is just the previous question. When $a=1$ we get the trivial group and otherwise we get a group isomorphic to the infinite cyclic group $\mathbb{Z}$.

$(g)\;$ Yes. Closed since $z^n=w^n=1\implies(zw)^n=1$, the identity is $z=1$ and inverses are found by $z^n=1\implies (z^{-1})^n=1$.

You should recognise it is isomorphic to the cyclic group $C_n=\mathbb{Z}/2$.

$(h)\;$ Yes, check it in a similar way to the previous part. (Are we bored yet?)

$(i)\;$ Yes. Or no. It depends…

if $R=1$, then yes. Closed since $|z|=|w|=1\implies|zw|=1$, identity $z=1$ and inverses $|z|=1\implies |z^{-1}|=1$.
However, these all break if $R\neq 1$.

$(j)\;$ No. If $R\leq 1$ then it doesn’t contain the identity. If $R>1$ then it isn’t closed.

Exercise 0.12 $\;$ Which of the following maps $f:\mathbb{C}^{\times}\rightarrow\mathbb{R}^{\times}$ are group homomorphisms?

$\;(a)\;$ $f(z)=z,\hspace{2cm}$ $\;(b)\;$ $f(z)=2|z|,\hspace{2cm}$ $\;(c)\;$ $f(z)=1/|z|,\hspace{2cm}$

$\;(d)\;$ $f(z)=1+|z|,\hspace{0.9cm}$ $\;(e)\;$ $f(z)=|z|^2,\hspace{2cm}$

$\;(f)\;$ $f(z)=1,\hspace{2cm}$ $\;(g)\;$ $f(z)=2$.

Solution

We just need to check $f(zw)=f(z)f(w)$ for all $z, w\in\mathbb{C}^\times$.

$(a)\;$ Yes, $|zw|=|z||w|$.

$(b)\;$ No, $2|zw|\neq 2|z|\cdot 2|w|$.

$(c)\;$ Yes, $\frac{1}{|zw|}=\frac{1}{|z|}\frac{1}{|w|}$.

$(b)\;$ No, $1+|zw|\neq |1+z|\cdot|1+w|$.

$(e)\;$ Yes, $|zw|^2=|z|^2|w|^2$.

$(f)\;$ Yes, $1=1\cdot 1$.

$(g)\;$ No, $2\neq 2\cdot 2$.

Exercise 0.13 $\;$ Find all isomorphisms from $\left(\mathbb{Z}/4,+\right)$ to $\left((\mathbb{Z}/5)^\times,\times\right)$

Solution

Both groups are cyclic of order $4$ and any group homomorphism $\varphi:\left(\mathbb{Z}/4,+\right)\to\left((\mathbb{Z}/5)^\times,\times\right)$ will be determined by the image of the generator $\overline{1}=1\bmod 4$ of $\left(\mathbb{Z}/4,+\right)$.

It will be an isomorphism precisely when $\varphi\left(\overline{1}\right)$ is a generator of $\left((\mathbb{Z}/5)^\times,\times\right)$. We can check these are $\overline{2}=2\bmod 5$ and $\overline{3}=3\bmod 5$ so there are just two possible isomorphisms. We can write them explicitly as \[\varphi(n \bmod 4)=2^n\bmod 5 \quad\text{and}\quad \varphi(n \bmod 4)=3^n\bmod 5.\]

Exercise 0.14 $\;$For an arbitrary group $G$, show that

$\;(a)\;$ the intersection of any number of subgroups of $G$ is also a subgroup,

$\;(b)\;$ the union of two subgroups is a group if and only if one of these subgroups is contained in the other.

Solution

$(a)\;$ Suppose we have subgroups $G_\alpha\subset G$ indexed by some set $\alpha\in\mathcal{I}$. If $a, b\in \bigcap_\alpha\,G_\alpha$, then $a, b\in G_\alpha$ for all $\alpha\in\mathcal{I}$. The identity $e$ of $G$ and $ab$ and $a^{-1}$ are in $G_\alpha$ for each $\alpha\in\mathcal{I}$. Hence they are also in the intersection. Associativity holds as it does in the overall group $G$.

$(b)\;$ Clearly, given subgroups $G_1, G_2\subset G$ with one of the subgroups contained in the other, then the union $G_1\cup G_2$ equals $G_1$ or $G_2$ so is a subgroup.

Conversely, suppose $G_1\cup G_2$ is a subgroup and one of the subgroups is not contained in the other. Then there must be $a\in G_1\setminus G_2$ and $b\in G_2\setminus G_1$. Now $a, b \in G_1\cup G_2\implies ab\in G_1\cup G_2$ and hence $ab$ is in either $G_1$ or $G_2$. However, $ab\in G_1\implies a^{-1}ab=b\in G_1$ and $ab\in G_2\implies abb^{-1}=a\in G_2$. Either way, we get a contradiction.

Exercise 0.15 $(\star)\;$ $\;$Given elements $a$ and $b$ in a group $G$, show that

$\;(a)\;$ $a$ and $b^{-1}ab$ have the same order,

$\;(b)\;$ $ab$ and $ba$ have the same order.

Solution

$(a)\;$ Use the trick that $\left(b^{-1}ab\right)^n=\left(b^{-1}ab\right)\left(b^{-1}ab\right)...\left(b^{-1}ab\right)=b^{-1}a^nb$.

If $a^n=e$ then $\left(b^{-1}ab\right)^n=b^{-1}eb=e$.
If $\left(b^{-1}ab\right)^n=e$ then $b^{-1}a^nb=e$ and so $a^n=e$.

$(b)\;$ Notice that $ab=b^{-1}(ba)b$ so we can use the previous part!

If this is hard to see, introduce some new letters: the previous part tells us $x$ and $y^{-1}xy$ have the same order. Now substitute $x=ba$ and $y=b$ so that $y^{-1}xy=ab$.

Exercise 0.16 $\;(a)\;$ Is the (additive) group $\mathbb{Z}/{15}$ cyclic?

$\;(b)\;$ Is the (multiplicative) group $(\mathbb{Z}/{16})^\times$ cyclic?

Solution

$(a)\;$ Yes! The group $\mathbb{Z}/n$ is cyclic for any $n$ since it is generated (additively) by $\overline{1}=1 \bmod n$.

$(b)\;$ No. There are $8$ invertible numbers $\bmod 16$ and \[\left(\mathbb{Z}/{16}\right)^\times=\left\{1,3,5,7,9,11,13,15\right\}.\]

So we need an element of order $8$. However, if $a\in\mathbb{Z}$ is odd, say $a=2b+1$, then \[a^2=4b(b+1)+1\equiv 1\bmod 8\quad\text{and}\quad a^4\equiv 1\bmod 16.\]

Actually, working a little bit harder, we can show that the group $\left(\mathbb{Z}/{16}\right)^\times$ is generated by the two elements $-1\bmod 16$ and $5\bmod 16$: \[\left(\mathbb{Z}/{16}\right)^\times=\langle\,\overline{-1}\,\rangle\times\langle\,\overline{5}\,\rangle\cong\mathbb{Z}/2\times\mathbb{Z}/4. \]

Exercise 0.17 $\;$Prove that the group $\mathbb{Q}$ with addition is not cyclic but is the union of its cyclic subgroups.

Solution

If $\mathbb{Q}$ is cyclic, then it has a (non-zero) generator $r\in\mathbb{Q}$ and all rational numbers are integer multiples, \[m\cdot r=\underbrace{r+r\,+\cdots+\,r}_\text{$m$ times}.\] But this misses lots of rational numbers, e.g. $r/2\in\mathbb{Q}$ is not of this form.

On the other hand, the cyclic subgroups of $\mathbb{Q}$ are $\{0\}$ and $\frac{1}{n}\mathbb{Z}=\left\{\frac{m}{n} \;|\; m\in\mathbb{Z}\right\}$ for each $n\in\mathbb{N}$. (Check this!)

We clearly have $\mathbb{Q}=\bigcup_{n\in\mathbb{N}}\;\frac{1}{n}\mathbb{Z}$.

Be careful with the notation here: the infinite group $\frac{1}{n}\mathbb{Z}$ is not the the finite group $\mathbb{Z}/n$.

More generally, one can show any group is the union of its cyclic subgroups $\displaystyle G=\bigcup_{g\in G}\;\langle\,g\,\rangle$.

Exercise 0.18 $\;$For any field $K$ and natural number $n$, show that the multiplicative subgroup $K^\times$ can not contain two distinct cyclic subgroups of order $n$.

Solution

Suppose that $G\subset K^\times$ is a cyclic subgroup of order $n$. Then $g^n=1$ in $K$ for each $g\in G$. However, the polynomial $x^n-1\in K[x]$ has degree $n$ so has at most $n$ roots. That means a generator of any cyclic group of order $n$, i.e. element of order $n$ in $K^\times$, is contained in $G$ and so generates $G$.

Exercise 0.19 $\;$Find all group homomorphisms between the following cyclic groups:

$\;(a)\;$ $\varphi:\mathbb{Z}/6\longrightarrow\mathbb{Z}/6,\hspace{2cm}$ $\;(b)\;$ $\varphi:\mathbb{Z}/6\longrightarrow\mathbb{Z}/{18},\hspace{2cm}$ $\;(c)\;$ $\varphi:\mathbb{Z}/{18}\longrightarrow\mathbb{Z}/6,$

$\;(d)\;$ $\varphi:\mathbb{Z}/{12}\longrightarrow\mathbb{Z}/{15},\hspace{1.5cm}$ $\;(e)\;$ $\varphi:\mathbb{Z}/6\longrightarrow\mathbb{Z}/{25}.$

Solution

Recall we write the group operation in the abelian group $\mathbb{Z}/m$ as addition so require \[\varphi(x+y)=\varphi(x)+\varphi(y).\]

We can use the fact that any homomorphism from a cyclic group to another group is uniquely determined by the image of a generator. So $\varphi:\mathbb{Z}/m\to\mathbb{Z}/n$ is determined by $a=\varphi(1\bmod m)$ and is of the form \[\begin{align*} \varphi_a\;:\; \mathbb{Z}/m\;\;\; &\longrightarrow \;\;\;\mathbb{Z}/n \\ k\bmod m &\longmapsto ka\bmod n \end{align*}\] To determine all possible group homomorphisms thus requires finding the list of possible values of $a$. To do this, remember $\varphi$ must send the identity element $0\bmod m$ in $\mathbb{Z}/m$ to the identity element $0\bmod n$ in $\mathbb{Z}/n$. Since $m\equiv 0\bmod m$, this means \[ma=m\varphi(1\bmod m)=\varphi(m\bmod m)=\varphi(0\bmod m)=0\bmod n \quad\text{in $\mathbb{Z}/n$}\] and we just need $a$ to satisfy $ma\equiv 0\bmod n$.

$(a)\;$ Here $a=\varphi(1\bmod 6)$ must be a solution of $6a\equiv 0\bmod 6$. This is true for any element $a\in\mathbb{Z}/6$ so there are exactly $6$ possible homomorphisms.

$(b)\;$ This time, $a=\varphi(1\bmod 6)$ is a solution of $6a\equiv 0\bmod 18$, i.e. any of the six values $0, 3, 6, 9, 12, 15 \bmod 18$.

$(c)\;$ Again, $a=\varphi(1\bmod 18)$ can be any element of $\mathbb{Z}/6$.

$(d)\;$ Now $a=\varphi(1\bmod 12)$ can be any of $0, 4, 8, 12 \bmod 15$.

$(e)\;$ Here $a=\varphi(1\bmod 6)$ must be $0 \bmod 25$ and there is only one homomorphism possible (the zero homomorphism).

Exercise 0.20 $\;$Prove that there are no surjective group homomorphisms $\mathbb{Q}\longrightarrow\mathbb{Z}$.

Solution

Suppose $\varphi:\mathbb{Q}\to\mathbb{Z}$ is a surjective homomorphism. Then $1\in\mathbb{Z}$ is in the image and so there is $r\in\mathbb{Q}$ such that $\varphi(r)=1\in\mathbb{Z}$. However, $r/2\in\mathbb{Q}$ so $m=\varphi(r/2)\in\mathbb{Z}$ must satisfy \[2m=2\varphi(r/2)=\varphi(r)=1\] which is impossible (as $m=1/2$ is not an integer!)

Exercise 0.21 $\;$Find all abelian groups of the following given orders:

$\;(a)\;$ $2,\hspace{4em}$ $\;(b)\;$ $6,\hspace{4em}$ $\;(c)\;$ $8,\hspace{4em}$ $\;(d)\;$ $12$,

$\;(e)\;$ $16,\hspace{3.5em}$ $\;(f)\;$ $24,\hspace{3.4em}$ $\;(g)\;$ $36.$

Solution

Recall the classification from The Fundamental Theorem of Finitely Generated Abelian Groups:

any non-trivial finite abelian group is (isomorphic to) exactly one of the following form \[\mathbb{Z}/d_1\,\times\,\mathbb{Z}/d_2\,\times\,...\,\times\,\mathbb{Z}/d_k\quad\text{where $d_1>1$ and $d_1\,|\,d_2\,|\,...\,|\,d_k$.}\] Using this we obtain the following lists of possibilities:

$(a)\;$ $\mathbb{Z}/2$,

$(b)\;$ $\mathbb{Z}/6$,

$(c)\;$ $\mathbb{Z}/8$, $\mathbb{Z}/2\times\mathbb{Z}/4$, $\mathbb{Z}/2\times\mathbb{Z}/2\times\mathbb{Z}/2$,

$(d)\;$ $\mathbb{Z}/12$, $\mathbb{Z}/2\times\mathbb{Z}/6$,

$(e)\;$ $\mathbb{Z}/16$, $\mathbb{Z}/2\times\mathbb{Z}/8$, $\mathbb{Z}/4\times\mathbb{Z}/4$, $\mathbb{Z}/2\times\mathbb{Z}/2\times\mathbb{Z}/4$, $\mathbb{Z}/2\times\mathbb{Z}/2\times\mathbb{Z}/2\times\mathbb{Z}/2$,

$(f)\;$ $\mathbb{Z}/24$, $\mathbb{Z}/2\times\mathbb{Z}/12$, $\mathbb{Z}/2\times\mathbb{Z}/2\times\mathbb{Z}/6$,

$(g)\;$ $\mathbb{Z}/36$, $\mathbb{Z}/2\times\mathbb{Z}/18$, $\mathbb{Z}/3\times\mathbb{Z}/12$, $\mathbb{Z}/6\times\mathbb{Z}/2$.

Exercise 0.22 $\;$How many subgroups are there

$\;(a)\;$ of order $2$ and $6$ in the non-cyclic abelian group of order $12$,

$\;(b)\;$ of order $3$ and $6$ in the non-cyclic abelian group of order $18$,

$\;(c)\;$ of order $5$ and $15$ in the non-cyclic abelian group of order $75$.

Solution

$(a)$ The (only) non-cyclic abelian group of order $12$ is \[G=\mathbb{Z}/2\times\mathbb{Z}/6\cong\mathbb{Z}/2\times\mathbb{Z}/2\times\mathbb{Z}/3.\]

There are three elements of order $2$ and therefore $3$ subgroups of order $2$.
There are two subgroups of order $6$, isomorphic to $\mathbb{Z}/6\cong\mathbb{Z}/2\times\mathbb{Z}/3$ (one for each $\mathbb{Z}/2$).

$(b)$ This time, \[G=\mathbb{Z}/3\times\mathbb{Z}/6\cong\mathbb{Z}/2\times\mathbb{Z}/3\times\mathbb{Z}/3.\]

There are four subgroups of order $3$.
There are four subgroups of order $6$.

$(c)$ Finally, \[G=\mathbb{Z}/5\times\mathbb{Z}/15\cong\mathbb{Z}/3\times\mathbb{Z}/5\times\mathbb{Z}/5.\]

There are six subgroups of order $5$.
There are six subgroups of order $15$.

Exercise 0.23 $\;(a)\;$ How many elements in $\mathbb{Z}/2\times\mathbb{Z}/4\times\mathbb{Z}/3$ have order $2$, $4$ and $6$.

$\;(b)\;$ How many elements in $\mathbb{Z}/2\times\mathbb{Z}/4\times\mathbb{Z}/4\times\mathbb{Z}/5$ have order $2$, $4$ and $5$.

Solution

One can use the multiplicative nature of the function \[A_m(G)=\left|\left\{ g\in G \;|\; \text{order of $g$ divides $m$}\right\} \right|\] as you did at the end of Algebra II. Alternatively, one can count them in an ad hoc way:

$(a)$ In $\mathbb{Z}/2\times\mathbb{Z}/4\times\mathbb{Z}/3$,

There are $3$ subgroups of order $2$ giving $3$ elements of order $2$.
Any subgroup of order $4$ is a subgroup of $\mathbb{Z}/2\times\mathbb{Z}/4$. In here there are $4$ elements of order $4$, namely $(0,1), (0,3), (1,1), (1,3)$.
To get an element of order $6$, we need an element of order $2$ in $\mathbb{Z}/2\times\mathbb{Z}/4$ combined with an element of order $3$ in $\mathbb{Z}/3$. There are thus $3\times 2$ choices giving $6$ elements of order $6$.

$(b)$ In $\mathbb{Z}/2\times\mathbb{Z}/4\times\mathbb{Z}/4\times\mathbb{Z}/5$

There are $7$ elements of order $2$.
There are $24$ elements of order $4$.
There are $4$ elements of order $5$.

$\;$

Exercise 0.24 $(\star)\;$ $\;$ As well as homomorphisms of groups, we can also sometimes consider groups of homomorphisms. Here’s an extended problem introducing the idea.

Suppose $A$ and $B$ are abelian groups (written using additive notation) and let $\Hom(A,B)$ denote the set of homomorphisms $\varphi:A\to B$. Given $\varphi,\varphi'\in\Hom(A,B)$, define the sum $\varphi+\varphi'\in\Hom(A,B)$ pointwise by \[(\varphi+\varphi')(a)=\varphi(a)+\varphi'(a) \quad\text{for all $a\in A$.} \]

$\;(a)$ Show that $\Hom(A,B)$ forms an abelian group under this addition.

$\;(b)$ Determine the following groups of homomorphisms:

$\;\;\;(i)\;$ $\Hom\left(\mathbb{Z}\,,\,\mathbb{Z}\right),\hspace{2.1cm}$ $\;\;\;(ii)\;$ $\Hom\left(\mathbb{Z}\,,\,\mathbb{Z}/n\right),\hspace{2cm}$ $\;\;\;(iii)\;$ $\Hom\left(\mathbb{Z}/n\,,\,\mathbb{Z}\right),$

$\;\;\;(iv)\;$ $\Hom\left(\mathbb{Z}/2\,,\,\mathbb{Z}/8\right),\hspace{0.9cm}$ $\;\;\;(v)\;$ $\Hom\left(\mathbb{Z}/2\,,\,\mathbb{Z}/7\right).$

$\;(c)$ Show that $\Hom(\cdot,\cdot)$ respects direct products, i.e. \[\begin{align*} \Hom\left(A_1\times A_2,B\right) &\cong \Hom(A_1,B)\times\Hom(A_2,B) \\ \text{and}\quad\Hom\left(A,B_1\times B_2\right) &\cong \Hom(A,B_1)\times\Hom(A,B_2). \end{align*}\] $\;(d)$ Determine the following groups of homomorphisms:

$\;\;\;(i)\;$ $\Hom\left(\mathbb{Z}/2\times\mathbb{Z}/2\,,\,\mathbb{Z}/8\right),\hspace{2cm}$ $\;\;\;(ii)\;$ $\Hom\left(\mathbb{Z}/2\times\mathbb{Z}/3\,,\,\mathbb{Z}/{30}\right).$

Solution

$(a)\;$ To show closure, i.e. that $\varphi+\varphi'\in\Hom(A,B)$, amounts to unraveling the brackets \[\begin{align*} (\varphi+\varphi')(a_1+a_2)&=\varphi(a_1+a_2)+\varphi'(a_1+a_2) \\ &=\varphi(a_1)+\varphi'(a_1)+\varphi(a_2)+\varphi'(a_2) \\ &=(\varphi+\varphi')(a_1)+(\varphi+\varphi')(a_2) \end{align*}\] for any $a_1,a_2\in A$. The identity element in $\Hom(A,B)$ is the zero-homomorphism $\varphi=0$, i.e. $\varphi(a)=0$ for any $a\in A$. Inverses are defined by $(-\varphi)(a)=-\varphi(a)$ for any $a\in A$. Finally, since $B$ is abelian we easily see \[(\varphi+\varphi')(a)=\varphi(a)+\varphi'(a)=\varphi'(a)+\varphi(a)=(\varphi'+\varphi)(a)\] so $\Hom(A,B)$ is an abelian group.

This is quite remarkable. We have a collection of objects (abelian groups) and the set of maps (group homomorphisms) between two of the objects is itself also one of the objects! Considering the structure of maps between objects rather than just objects themselves is a very fruitful idea in advanced mathematics.

$(b)(i)\;$ Since $\mathbb{Z}$ is cyclic, any element of $\Hom\left(\mathbb{Z}\,,\,\mathbb{Z}\right)$ is determined by the image of the (additive) generator $1\in\mathbb{Z}$. In particular, it is of the form \[\begin{align*} \varphi_a\;:\; \mathbb{Z} &\longrightarrow \mathbb{Z} \\ r &\longmapsto ra \end{align*}\] where $a=\varphi(1)$ is an arbitrary element of $\mathbb{Z}$. Also, notice $\varphi_a+\varphi_b=\varphi_{a+b}$ and thus $\Hom\left(\mathbb{Z}\,,\,\mathbb{Z}\right)\cong\mathbb{Z}$.

$(ii)\;$ Similarly, any homomorphism $\varphi:\mathbb{Z}\to\mathbb{Z}/n$ is determined by the image $a=\varphi(1)\in\mathbb{Z}/n$ and is of the form \[\begin{align*} \varphi_a\;:\; \mathbb{Z} &\longrightarrow \mathbb{Z}/n \\ r &\longmapsto ra\bmod n \end{align*}\] Again, we have $\varphi_a+\varphi_b=\varphi_{a+b}$ (where subscripts are added mod $n$) and $\Hom\left(\mathbb{Z}\,,\,\mathbb{Z}/n\right)\cong\mathbb{Z}/n$.

$(iii)\;$ As $\mathbb{Z}/n$ is cyclic, any $\varphi\in\Hom\left(\mathbb{Z}/n\,,\,\mathbb{Z}\right)$ is determined by the image $a=\varphi(1\bmod n)\in\mathbb{Z}$. However, this time $a$ is not arbitrary: we have \[na=n\varphi(1\bmod n)=\varphi(n\bmod n)=\varphi(0)=0\in\mathbb{Z}\] and so $a=0$ since $n\neq 0$. That means $\varphi$ is the zero homomorphism and $\Hom\left(\mathbb{Z}/n\,,\,\mathbb{Z}\right)=\{0\}$ is the trivial group.

Less explicitly but actually easier… the image of a homomorphism $\varphi:A\to B$ is a subgroup of $B$. Now if $A$ is finite then so is $\im(\varphi)$. But the only finite subgroup of $\mathbb{Z}$ is the trivial group! In particular, we find that $\Hom(A,\mathbb{Z})=\{0\}$ for any finite group $A$.

$(iv)\;$ This time, homomorphisms are of the form \[\begin{align*} \varphi_a\;:\; \mathbb{Z}/2 &\longrightarrow \mathbb{Z}/8 \\ r\bmod 2 &\longmapsto ra\bmod 8 \end{align*}\] where $a=\varphi_a(1\bmod 2)\in\mathbb{Z}/8$. But $2a=\varphi_a(2\bmod 2)=0\bmod 8$ means $a=0$ or $4\bmod 8$. We obtain $\Hom\left(\mathbb{Z}/2\,,\,\mathbb{Z}/8\right)\cong\mathbb{Z}/2$.

$(v)\;$ The image $a=\varphi(1\bmod 2)$ has order dividing $2$ since $2a=\varphi(2\bmod 2)=\varphi(0)=0$. However, $a$ is an element of $\mathbb{Z}/7$ so it also has order dividing $7$. That means $a$ has order $1$ and must equal $0$. We’ve found $\Hom(\mathbb{Z}/2\,,\,\mathbb{Z}/7)\cong\{0\}$ is the trivial group.

More generally, similar considerations show $\Hom(\mathbb{Z}/m\,,\,\mathbb{Z}/n)\cong\mathbb{Z}/d$ where $d=\gcd(m,n)$.

$(c)\;$ A direct product $A_1\times A_2$ contains copies of $A_1\cong A_1\times\{0\}$ and $A_2\cong \{0\}\times A_2$ as subgroups. Restricting a given homomorphism $\varphi:A_1\times A_2\to B$ to these gives homomorphisms \[\begin{align*} \varphi|_{A_1}\;:\; A_1 &\longrightarrow B \qquad\qquad\text{and}\quad & \varphi|_{A_2}\;:\; A_2 &\longrightarrow B \\ a_1 &\longmapsto \varphi(a_1,0) & a_2 &\longmapsto \varphi(0,a_2) \end{align*}\] and we can recombine these to define a map \[\begin{align*} F\;:\; \Hom\left(A_1\times A_2,B\right) &\longrightarrow \Hom(A_1,B)\times\Hom(A_2,B) \\ \varphi &\longmapsto \left(\varphi|_{A_1},\varphi|_{A_2}\right) \end{align*}\] We claim, that this is a group isomorphism. Firstly, if $\varphi, \varphi'\in\Hom(A_1\times A_2,B)$ then use \[\left(\varphi\pm\varphi'\right)|_{A_i}=\varphi|_{A_i}\pm\varphi'|_{A_i}\qquad\text{for $i=1,2$}\] to verify $F$ is a group homomorphism. Secondly, given $\varphi\in\ker(F)$, we have $\varphi|_{A_1}=\varphi|_{A_2}=0$. But then for any $(a_1,a_2)\in A_1\times A_2$, one sees \[\varphi\left((a_1,a_2)\right)=\varphi\left((a_1,0)+(0,a_2)\right)=\varphi|_{A_1}(a_1)+\varphi|_{A_2}(a_2)=0+0=0.\] In other words, $\varphi$ is the zero-homomorphism and $\ker(F)=\{0\}$. This proves $F$ is injective.

Now suppose $\varphi_1\in\Hom(A_1,B)$ and $\varphi_2\in\Hom(A_2,B)$. Define $\varphi\in\Hom(A_1\times A_2,B)$ by \[\varphi\left((a_1,a_2)\right)=\varphi_1(a_1)+\varphi_2(a_2)\quad\text{for any $(a_1,a_2)\in A_1\times A_2$.}\] Then $\varphi|_{A_1}(a_1)=\varphi\left((a_1,a_2)\right)=\varphi_1(a_1)$ for any $a_1\in A_1$ so $\varphi|_{A_1}=\varphi_1$. In a similar fashion, we get $\varphi|_{A_2}=\varphi_2$ and so $F(\varphi)=(\varphi_1,\varphi_2)$. This proves $F$ is surjective.

The proof of the other isomorphism $\Hom\left(A,B_1\times B_2\right)\cong\Hom(A,B_1)\times\Hom(A,B_2)$ is very similar and left for the enthusiastic reader!

$(d)(i)\;$ Using the above, we have $\Hom\left(\mathbb{Z}/2\times\mathbb{Z}/2\,,\,\mathbb{Z}/8\right)\cong\mathbb{Z}/2\times\mathbb{Z}/2$.

$(ii)\;$ In an entirely similar way, $\Hom\left(\mathbb{Z}/2\times\mathbb{Z}/3\,,\,\mathbb{Z}/{30}\right)\cong\mathbb{Z}/2\times\mathbb{Z}/3.$