cyclic groups

**nataliemarie** · February 24th 2010, 04:16 PM

Ok so the question is supposing that a,b are in Group G, the problem doesn't state that the group is cyclic or anything. But a,b in G and a has an odd order. then if
aba-1 = b-1 then b^2 = e. I did it at first but unfortunately I assumed the group is commutative which it isn't. Any hints on how to solve it if its not commutative. I was trying multiplying both sides of the equation by a, but I'm not getting anywhere, and am in desperate need of ideas. Thanks!!

**tonio** · February 24th 2010, 07:41 PM

Originally Posted by nataliemarie

Ok so the question is supposing that a,b are in Group G, the problem doesn't state that the group is cyclic or anything. But a,b in G and a has an odd order. then if
aba-1 = b-1 then b^2 = e. I did it at first but unfortunately I assumed the group is commutative which it isn't. Any hints on how to solve it if its not commutative. I was trying multiplying both sides of the equation by a, but I'm not getting anywhere, and am in desperate need of ideas. Thanks!!

(***) $aba^{-1}=b^{-1}$ . Take inverses in both sides:

$ab^{-1}a^{-1}=b$ , and now use (***) above in the left side:

$a(aba^{-1})a^{-1}=b\Longrightarrow a^2ba^{-2}=b$ . Again, take inverses in both sides:

$a^2b^{-1}a^{-2}=b^{-1}$ , and again use (***) in the left side:

$a^3ba^{-3}=b^{-1}$ ....

Can you see the pattern here? Well, now follow it and use the fact that the order of $a$ is odd.

Tonio

**nataliemarie** · February 25th 2010, 09:07 AM

Ok i did it. Am I allowed to make a^3 = identity. I know the identity is odd, and i know any multiple of the identity is e, and the identity should be the smallest possible number that will give you the identity.

**tonio** · February 25th 2010, 07:29 PM

Originally Posted by nataliemarie

Ok i did it. Am I allowed to make a^3 = identity. I know the identity is odd, and i know any multiple of the identity is e, and the identity should be the smallest possible number that will give you the identity.

No, of course you're not allowed to put $a^3=1$ : why would you?!? Follow the pattern knowing that $a^{2k-1}=1$ for some $k\in\mathbb{N}$ .

All what you wrote after " I know the..." makes no sense at all. Please do check your notes.

Tonio

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