Current Events > Math gurus, I need help!

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FLOUR 09/07/21 6:51:39 PM #1:	Let G be a group of finite order. I need to prove that the function f(x)=x^2 acts as an automorphism on G if and only if G is abelian with odd order. --- You don't want to play cards with me because I'll cheat. You wanna play 21? I've got 22! <cite>FLOUR posted [p:957828897] in Math gurus... [t:79650573]...</cite> <quote>Let G be a group of finite order. I need to prove that the function f(x)=x^2 acts as an automorphism on G if and only if G is abelian with odd order. </quote> ... Copied to Clipboard!
nothanks1 09/07/21 6:52:45 PM #2:	https://math.stackexchange.com/questions/738500/let-g-be-an-abelian-group-with-odd-order-show-that-varphi-g-to-g-such-t/2712162 <cite>nothanks1 posted [p:957828930] in Math gurus... [t:79650573]...</cite> <quote>https://math.stackexchange.com/questions/738500/let-g-be-an-abelian-group-with-odd-order-show-that-varphi-g-to-g-such-t/2712162</quote> ... Copied to Clipboard!
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DespondentDeity 09/07/21 6:53:55 PM #4:	We got a special report from Alex Jones here: math is gay, I think math is kinda gay --- The web of destiny carries your blood and soul back to the Genesis of my life form. I'm softer than a daisy, if you cut me I'll bleed pink <cite>DespondentDeity posted [p:957828968] in Math gurus... [t:79650573]...</cite> <quote>We got a special report from Alex Jones here: math is gay, I think math is kinda gay </quote> ... Copied to Clipboard!
trappedunderice 09/07/21 6:54:16 PM #5:	Well you see Since G G has odd order, that means G G is finite. So can be injective if and only if it is surjective. To formally prove surjectivity, argue by contradiction. Suppose there is a yG yG such that xG xG, (x)y (x)y. That implies there are two distinct x 1 ,x 2 x1,x2 such that (x 1 )=(x 2 ) (x1)=(x2) by injectivity. <cite>trappedunderice posted [p:957828971] in Math gurus... [t:79650573]...</cite> <quote>Well you see Since G G has odd order, that means G G is finite. So can be injective if and only if it is surjective. To formally prove surjectivity, argue by contradiction. Suppose there is a yG yG such that xG xG, (x)y (x)y. That implies there are two distinct x 1 ,x 2 x1,x2 such that (x 1 )=(x 2 ) (x1)=(x2) by injectivity. </quote> ... Copied to Clipboard!
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