Determine whether the subsets H is a sugroup of G, if it is show if its abelian, if not a subgroup, show $ab^{-1}$ is not in H for a,b in H.

Question

1)$G=D_n$ be a dihedral group of order $2n$, $n\ge 3$. Let H = $[a\in G|a^2 = e]$.

My answer: H = {$e,s,rs,...,r^{n-1}s$} so ab is in H for all a,b in H. So H is a subgroup, but H is not abelian since $r^as^b = s^{a+b}$ does not equal $s^{a-b}= s^br^a \:mod(n)$

2)G = $GL(2,\mathbb{R})$ and let H = {$A=\begin{pmatrix} u& -v \\ v &u \end{pmatrix}|v^2+u^2=1$}

My answer: H is obviously a subgroup since Det(AB)=Det(A)Det(B)=1 for all A,B in G, but its not abelian since n = 2.

3)G is an abelian group and H = {$a \in G|a\: has \: finite \: order$}. Im not quite sure how to approach this one.

Answer 1

For 3),

Recall that in an abelian group, $(ab)^n=a^nb^n$. Clearly the identity of $G$ has finite order.

If $a$ has finite order $m$, and $b$ has finite order $n$, what can you say about $(ab)^{mn}$? (Or $(ab)^{\operatorname{lcm}(m,n)}$ even.)

If $a$ has finite order $n$, can you show that $a^{-1}$ also has order $n$?

These three steps should all lead you to conclude that $H$ is indeed a subgroup of $G$, when $G$ is abelian.

Answer 2

In an abelian group $G$, if the order of an element $x$ is denoted o($x$) then there is an onto homomorphism $\eta:\left<a\right>\times\left<b\right>\longrightarrow\left<a,b\right>$ where $\left< S\right> $ denotes the subgroup generated by the elements of $S\subseteq G$. This implies o($a b^{-1}$) divides o($a$) o($b$).