Subgroup proof is this right?

sfspitfire23 · #1 $Old$ November 3rd, 2009, 10:23 AM

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 $a^{-1} \in H$ .

aliceinwonderland · #2 $Old$ November 3rd, 2009, 09:10 PM

Quote:

Originally Posted by sfspitfire23 $View Post$

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 $S_3$ .

redsoxfan325 · #3 $Old$ November 3rd, 2009, 10:35 PM

Quote:

Originally Posted by sfspitfire23 $View Post$

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 $a^{-1} \in H$ .

Problem: Let $H$ be any group. Define $S=\{a\in H~|~a^2=1\}$ . Prove that $S$ is a subgroup of $H$ .

Proof (of inverse): Given $a^{-1}\in H$ , we have to show that $a^{-1}\in S$ if $a\in S$ .

$(a^{-1})^2=a^2(a^{-1})^2=1$ so $a^{-1}\in S$

$\square$