## Algebra Solutions: Hungerford I.6

2)(a) $S_n$ is generated by the $n-1$ transpositions: $(12),(13),...,(1n)$
Proof-
Let $\alpha \in S_n$. We can write $\alpha = (i_1i_2)(i_3i_4) \cdots (i_{r-1}i_r)$, where $2 \leq i_r \leq n$. Note that each $(i_ki_{k+1}) = (1i_k)(1i_{k+1})(1i_k)$. Using this we can write $\alpha = (1i_1)(1i_2)\cdots (1i_{r-1})(1i_r)(1i_{r-1} \in \langle (12),(13),...,(1n)\rangle$. Thus any permutation in $S_n$ can be written as a product of the given transpositions.

(b) $S_n$ is generated by the $n-1$ transpositions $(12),(23),..., ((n-1) n)$
Proof-
Let $\alpha \in S_n$. From part a we can write $\alpha = (1i_1)(1i_2)\cdots (1i_r)$ where $2 \leq i_r \leq n$. Now consider $(1i_k)$. If $i_k = 2$ then leave $(1i_k)$ alone. If $i_k > 2$ then write $(1i_k) = (1i_k-1)(i_k-1i_k)(1i_k-1)$. Repeat this process on each transposition of the form $(1i_k-j)$ in the composition of $\alpha$for each $j \geq$ as long as $i_k-j>2$. Doing this gives $\alpha$ written as a product of transpositions of the form $(i_k-1i_k)$ so that we are done.

12) The center of the group $D_n$ is ${e}$ if n is odd and $\mathbb{Z}_2$ if n is even.
Proof-
First we need to know all the normal subgroups in $D_n$. If n is odd, the nontrivial subgroups are the group of rotations, and all the subgroups of the group of rotations. If n is even we have two more normal subgroups, each isomorphic to $D_{\frac{n}{2}}$.
Let H be a normal subgroup. We have two cases, either H consists only of powers of the element a, or there is some $a^ib \in H$. Suppose that $H = \langle a^k \rangle$ for some k dividing n. Then:
$a^i(a^k)^ra^{-i} = (a^k)^r \in \langle a^k \rangle$
$b(a^k)^rb = b^2(a^{-2kr})=((a^k)^r)^{-1} \in \langle a^k \rangle$ because of the relation $ba^r = a^{-r}b$. And finally,
$a^ib(a^k)^ra^ib = a^i(a^{-kr-i)}=a^{-kr} \in \langle a^k \rangle$. Thus the subgroup is normal.
Now instead suppose there exists some $a^kb \in H$. Then we have $a^{k+2}b = aa^k(ab) = a(a^kb) a^{-1} \in H$ since H is normal. This gives that $(a^{k+2}b)(a^kb) = a^{k+2} a^{-k} = a^2\in H$. This tells us that $a^2$ is a generator of H.
Next, $a^{k-2l}b = (a^2)^{-l}(a^kb) \in H \forall l \in \mathbb{Z}$. If $k= 2m$ is even let $l = m \Rightarrow b\in H$ and if $k = 2m+1$ is odd let $l = m \Rightarrow ab\in H$. Therefore either b is a generator or ab is a generator. So that $H = \langle a^2,b \rangle$ or $H = \langle a^2,ab \rangle$.
Now that we know all the normal subgroups we can find the center, because it must be a normal subgroup. Recall the center is the set of all elements which can commute with any other element.
First we check $H = \langle a^2,b \rangle$. We have $ab = ba \Rightarrow a^{-1} = a \Rightarrow a^2 = e$, but n is larger than 2 so this is a contradiction. Thus this is not the center.
Next we check $H = \langle a^2,ab \rangle$. We have that:
$a^2(ab) = (ab)a^2$
$a^3b = a^{-1}b \Rightarrow a^3 = a^{-1} \Rightarrow a^4 = e$. But for n = 4, $(ab)b = b(ab) \Rightarrow a = a^{-1} \Rightarrow a^2 = e$ again a contradiction so this is not the center.
Lastly, assume that $\langle a^k \rangle$ is the center. Then we have $ba^k = a^kb = ba^{-k} \Rightarrow a^{2k} = e \Rightarrow 2k = n \Rightarrow k = \frac{n}{2}$. Thus the center can only be $\langle a^{\frac{n}{2}} \rangle$ which the same as $\mathbb{Z}_2$ but only when n is even. If n is odd then the center is just the identity.