Prerequisites

Solution: We have to verify the group axioms. Firstly, for any $x,y,z\in G$,

showing that $\ast_{g}$ is associative (here we used that we know that $\cdot$ is associative). Next we verify that $g^{-1}$ is an identity for $\ast_{g}$:

Finally, we show that every element has an inverse: for any $x\in G$,

(it follows that $g^{-1}x^{-1}g^{-1}$ is also a left inverse of $x$).

Solution: Recall that the order of any element and any subgroup must divide the order of the group. Since $D_{4}=\{e,s,r,sr,r^{2},sr^{2},r^{3},sr^{3}\}$, there are eight elements, hence any subgroup must have order $1,2,4$, or $8$. The subgroup of order one is $\{e\}$, and the order eight subgroup is $D_{4}$ itself. The order two subgroups are all of the form $\{e,x\},$ where $x^{2}=e$. Going through the elements one by one, we see that the possibilities are $\{e,s\}$, $\{e,r^{2}\}$, $\{e,sr\}$, $\{e,sr^{2}\}$, $\{e,sr^{3}\}$. Finally, we search for the order four subgroups. Clearly $\{e,r,r^{2},r^{3}\}$ is a cyclic subgroup of order four. Noting that $r$ and $r^{3}$ are the only group elements of order four, the other subgroups of order four must be of the form $\{e,x,y,xy\}$, where $x$ and $y$ are two commuting elements of order two, and hence $xy$ is also of order two. Systematically multiplying together the order two elements gives the subgroups $\{e,s,r^{2},sr^{2}\}$, $\{e,sr,r^{2},sr^{3}\}$, and no other possibilities.

Solution: Recall that $Q_{8}=\{\pm 1,\pm i,\pm j,\pm k|i^{2}=j^{2}=k^{2}=ijk=-1\}$. The systematic way to find conjugacy classes is to go through the group element by element, repeatedly conjugating by the generators until all elements have been assigned to a conjugacy class. The singleton $\{1\}$ is a conjugacy class. $-1\in Z(Q_{8})$, so it too is its own conjugacy class. Then taking the element $i$ and conjugating by the generators gives $ii(-i)=i$, $ji(-j)=kj=-i$, and $ki(-k)=-jk=-i$ Since $-i=(-1)i$, and $-1\in Z(Q_{8})$, conjugating $-i$ by the generators gives no new elements, hence $\{\pm i\}$ is a conjugacy class. By symmetry in $i,j,k$, the remaining conjugacy classes are $\{\pm j\}$ and $\{\pm k\}$.

There are therefore 5 conjugacy classes of $Q_{8}$: $\{1\}$, $\{-1\}$, $\{\pm i\}$, $\{\pm j\}$, and $\{\pm k\}$.

Solution: We proceed as in Problem 3: $D_{4}=\{e,s,r,sr,r^{2},sr^{2},r^{3},sr^{3}\}$, with generators $r$ and $s$. The singleton $\{e\}$ is a conjugacy class. Conjugating $s$ by the generators gives $sss=s$, and $rsr^{3}=sr^{2}$, so $sr^{2}$ is in the same conjugacy class as $s$. Now conjugating $sr^{2}$ by the generators yields $ssr^{2}s=sr^{2}$, and $rsr^{2}r^{3}=s$, hence $\{s,sr^{2}\}$ is a conjugacy class. Conjugating $r$ by the generators: $rrr^{3}=r$, $srs=ssr^{3}=r^{3}$. Since all elements of a conjugacy class have the same order, and $r$ and $r^{3}$ are the only order four elements, no other elements can be in the conjugacy class $\{r,r^{3}\}$. Starting with the element $sr$: $ssrs=sr^{3}$, and $rsrr^{3}=sr^{3}$. Now conjugating $sr^{3}$: $ssr^{3}s=sr,$ and $rsr^{3}r^{3}=sr$, giving the conjugacy class $\{sr,sr^{3}\}$. All that remains is $\{r^{2}\}$,

i.e. the conjugacy classes of $D_{4}$ are as follows: $\{e\}$, $\{r^{2}\}$, $\{s,sr^{2}\}$, $\{r,r^{3}\}$, and $\{sr,sr^{3}\}$.

Solution: $\{e\}$ is of course a conjugacy class. The other elements are all of the form $r^{j}$ or $sr^{j}$. Since $\langle r\rangle$ is cyclic (and hence Abelian), conjugating $r^{j}$ with an element $r^{k}$ just returns $r^{j}$. On the other hand, conjugating $r^{j}$ with $sr^{k}$ gives

This shows that $\{r^{j},r^{n-j}\}$ is a conjugacy class for $1\leq j\leq\frac{n}{2}$ (Note that if $n$ is even, then $\{r^{n/2}\}$ is its own conjugacy class.) Now taking an arbitrary $sr^{j}$ and conjugating with $r$ gives

Since $sr^{j-2}=sr^{j+n-2}$, we see that if $n$ is odd, given any $sr^{j_{1}}$ and $sr^{j_{2}}$, we can find $k$ s.t. $r^{k}sr^{j_{1}}r^{-k}=sr^{j_{2}}$.

Thus, for odd $n$ the conjugacy classes are $\{e\}$, $\{r,r^{n-1}\}$, $\{r^{2},r^{n-2}\},\ldots,$ $\{r^{(n-1)/2},r^{(n+1)/2}\}$, and $\{sr^{k}\,|\,k=0,\ldots,n-1\}$.

On the other hand, if $n$ is even, then we see that repeatedly conjugating $sr^{j}$ with $r$ returns all elements $sr^{k}$ with $j\equiv_{2}k$. We also check that conjugating with $s$ has the same behaviour:

again since $n$ is even, $j$ and $n-j$ have the same parity mod $2$. The conjugacy classes for even $n$ are therefore $\{e\}$, $\{r^{n/2}\},$ $\{r,r^{n-1}\}$, $\{r^{2},r^{n-2}\},\ldots,$ $\{r^{n/2-1},r^{n/2+1}\}$, $\{sr^{k}\,|\,k\equiv_{2}0\}$, $\{sr^{k}\,|\,k\equiv_{2}1\}$.

Solution: Suppose $g,h\in\mathrm{Stab}_{G}(x)$, then

Thus $gh\in\mathrm{Stab}_{G}(x)$. Also note that, for any $g\in\mathrm{Stab}_{G}(x)$,

and thus $g^{-1}\in\mathrm{Stab}_{G}(x)$. As $\mathrm{Stab}_{G}(x)$ is clearly nonempty ($e\in\mathrm{Stab}_{G}(x)$) it is a subgroup of $G$.

Solution: Labelling the vertices of the triangle counterclockwise from (1,0), we see that $r$ moves $1$ to the position of $2$, $2$ to $3$’s spot, and $3$ back to $1$’s place. $s$ on the other hand leaves $1$ fixed and swaps $2$ and $3$. This suggests we should try to make a homomorphism $\varphi:D_{3}\rightarrow S_{3}$ by defining

In order for this to define a homomorphism, we simply need to verify that $\varphi(r)$ and $\varphi(s)$ satisfy the same relations as $r$ and $s$. $\varphi(s)$ is a two-cycle, and $\varphi(r)$ is a three-cycle, so $\mathrm{Id}=\varphi(s)^{2}=\varphi(r)^{3}$. It remains to check the relation $\varphi(s)\varphi(r)=\varphi(r)^{-1}\varphi(s)$. On the one hand,

while using that $(1\,2\,3)^{-1}=(1\,3\,2)$ gives

hence $\varphi$ is a homomorphism. In order to show that it is an isomorphism, we compute $\ker\varphi$. All group elements are of the form $s^{k}r^{j}$, where $k=0$ or $1$, and $j=0,1,2$. Looking at the elements $r^{j}$, if $r^{j}\in\ker\varphi$, then $\varphi(r^{j})=(1\,2\,3)^{j}=\mathrm{Id}$ only holds if $j=0$. For the elements $s,sr,sr^{2}$, we have $\varphi(s)=(2\,3)$, $\varphi(sr)=(1\,3),$ and $\varphi(sr^{2})=(2\,3)(1\,3\,2)=(1\,2)$, hence $\ker\varphi=\{e\}$. Then by the isomorphism theorem, $D_{3}\cong\mathrm{im}(\varphi)/\ker(\varphi)=\mathrm{im}(\varphi)$. However, since $|D_{3}|=|S_{3}|=6$, we must have $\mathrm{im}(\varphi)=S_{3}$.

Solution: We will show this by constructing a group homomorphism and then showing it is an isomorphism. Consider the map

This is a groups homomorphism as all relations are preserved by $\phi$: $s^{2}=e$, $(st)^{3}=e$ and $s(st)=t=(ts)s=(st)^{2}s$. As both groups have order $6$ and the kernel is trivial we have a bijection. Thus the groups are isomorphic

Solution: Label the elements of $G:$ $g_{1},\,\ldots g_{|G|}$. For each group element $h\in G$, we define a permutation $\sigma_{h}\in S_{|G|}$ by the rule $hg_{i}=g_{\sigma_{h}(i)}$. Since the map $g\mapsto hg$ is a bijection on $G$, $h\mapsto\sigma_{h}$ is a well-defined map from $G$ to $S_{|G|}$. We claim that it is a homomorphism:

hence $\sigma_{gh}=\sigma_{g}\circ\sigma_{h}$ for all $g,h\in G$. We also claim that $\ker(g\mapsto\sigma_{g})=\{e\}$: note that $\sigma_{g}(i)=i\Leftrightarrow gg_{i}=g_{i}\Leftrightarrow g=e$. Thus $G\cong\mathrm{im}(g\mapsto\sigma_{G})<S_{|G|}$.

Solution:

Solution: