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aliceinwonderland · #2 November 3rd, 2009, 08:10 PM

Quote:

Originally Posted by sfspitfire23

show that a^2=1 is a subgroup of h is my inverse right?

For the inverse, $(a^{-1})^2=1$ . So, $a^{-1}a^{-1}=1$ and thus $a^{-1} \in h$ .

Assume H is an abelian group. Let K be a subset of H such that $K=\{a \in H | a^2=1\}$ . We show that K is a subgroup of H. It suffices to show that whenever x and y are in K, then $xy^{-1}$ is also in K (link).

Since H is an abelian group, we have ${(xy^{-1})}^2=xy^{-1}xy^{-1}=xxy^{-1}y^{-1}=1$ . Thus K is a subgroup of H.

Anyhow I don't think K is necessarily a subgroup of H if H is a non-abelian group. Take an example of $S_3$ .

The following users thank aliceinwonderland for this useful post:
sfspitfire23
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