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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.

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Did you mean "subgroup" where you wrote "subspace"? –  DonAntonio Jan 28 '13 at 17:56
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Could you please make the question self-contained, as the title seems to end with a sentence fragment. –  Tobias Kildetoft Jan 28 '13 at 17:58
    
@Tobias hopefully that clarifies things –  bobdylan Jan 28 '13 at 18:05
    
@DonAntonio ohh yes, my mistake –  bobdylan Jan 28 '13 at 18:06
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@bobdylan It is NOT foolishness. The group $H$ is called the orthogonal group as its elements can be written as 2 by 2 matrices of the form $\left[ \begin{array}[2,2] {} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array}\right]$ and that is a rotation through an angle of $\theta$. This beautiful piece of linear algebra often slips through the cracks of high school and college curricula. –  Barbara Osofsky Jan 28 '13 at 19:42

2 Answers 2

up vote 1 down vote accepted

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.

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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($b$).

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