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I'm still working through Mac Lane and Birkhoff and I have another question. The question is: In a group $G$, show that $$ (g_1 g_2)^m = g_1^m g_2^m $$ for all integers $m$ if and only if $g_1 g_2 = g_2 g_1$.

It's pretty clear that if we assume that $g_1$ and $g_2$ then the above is true. However, I'm getting stuck trying show in the other direction. What I've been trying is to start with $$ (g_1 g_2)^m = g_1 g_1 g_1 \cdots g_2 g_2 g_2 $$ and then somehow cancel terms to arrive at the desired result of $g_1 g_2$ on the LHS and $g_2 g_1$ on the RHS (as this is pretty clear and works nicely when $m =2$). Is this the way to go about it, or is there another way?

raynea
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2 Answers2

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HINT

For $m=2$ we get $$ abab = (ab)^2 = a^2 b^2 = aabb \iff abab = aabb $$ premultiply by $a^{-1}$ and postmultiply by $b^{-1}$ to get $ba=ab$.

gt6989b
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  • Thanks. I was able to work through the case when $m=2$, but I could not get this same technique to work for generic $m$. For example, when you multiply by $a^{-1}$ on the left and $b^{-1}$, the result is $bababa\cdots a = a a \cdots b$ (now with $m-1$ terms on each side). From here, if you multiply by $b^{-1}$ on the left, then it does cancel the $b$ on the LHS, but you end up with $b^{-1} a a \cdots b$ on the RHS. This is where I get stuck. – raynea Jun 10 '20 at 13:10
  • @raynea you don't need to do it for each $m$, -- since you know it for all $m$, in particular, $m=2$ holds – gt6989b Jun 10 '20 at 14:12
  • Ah right! Thank you! – raynea Jun 10 '20 at 16:50
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If you take $m = -1$, we get that $$(g_1g_2)^{-1} = g_1^{-1}g_2^{-1} = (g_2g_1)^{-1}$$(Check the last equality). Since a group element times its inverse is the identity (denote it by $e$), we have $$e = (g_1g_2)(g_1g_2)^{-1} = (g_1g_2)(g_2g_1)^{-1}$$ Thus, by mutiplying by $(g_2g_1)$ on the right, we conclude that $(g_1g_2) = (g_2g_1)$.

pallatinus
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