Let $G_1=G_2=\dots=G_n$, and let $G=G_1 \times G_2 \times \cdots G_n$. For a permutation $\pi \in S_n$ define $$\varphi_\pi:G \to G$$ according to the rule $$\varphi_\pi(g_1,\dots,g_n)=(g_{\pi^{-1}(1)},\dots,g_{\pi^{-1}(n)}). $$ I'm required to prove that $\Phi:\pi \mapsto \varphi_\pi $ is a homomorphism. Here is my (failed) attempt:

We see how $\Phi(\pi_1 \pi_2)$ acts on a general element $g=(g_1,\dots,g_n)$ of the product:

$$ \begin{align*} \Phi(\pi_1 \pi_2)(g) &= \varphi_{\pi_1 \pi_2}(g_1,\dots,g_n) \\ &= (g_{(\pi_1 \pi_2)^{-1} (1)},\dots,g_{(\pi_1 \pi_2)^{-1}(n)})\\ &= (g_{\pi_2^{-1}(\pi_1^{-1}(1))},\dots,g_{\pi_2^{-1}(\pi_1^{-1}(n))}) \\ &= \varphi_{\pi_2}(g_{\pi_1^{-1}(1)},\dots,g_{\pi_1^{-1}(n)}) \\ &= \varphi_{\pi_2} \circ \varphi_{\pi_1}(g) \\ &= \left(\Phi(\pi_2) \circ \Phi(\pi_1) \right)(g) \end{align*}$$

I seem to have gotten $\Phi(\pi_1 \pi_2)=\Phi(\pi_2) \Phi(\pi_1)$ rather than the usual homomorphism law. Where have I gone wrong?


The way I see it, applying $\varphi_\pi$ to an $n$-tuple means, applying the inverse permutation $\pi^{-1}$ to each of the subscripts. Could this be my problem?

  • $\begingroup$ Yes, that is your problem. I'm fairly sure I have addressed a nearly identical question before. Give me a moment. $\endgroup$ – Jyrki Lahtonen Dec 28 '14 at 12:21
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    $\begingroup$ The elements of $G$ can be thought of as functions from the set $I=\{1,2,\ldots,n\}$ to the group $G_1$. If $f$ is such a function, and $\pi\in S_n$ is a permutation, then $\phi_\pi(f)=f\circ \pi^{-1}$. Thus $$\phi_{\pi_1}(\phi_{\pi_2}(f))=\phi_{\pi_1}(f\circ\pi_2^{-1})=f\circ\pi_2^{-1}\circ\pi_1^{-1}=f\circ(\pi_1\pi_2)^{-1}$$ et cetera. See for example here for a similar argument. I won't let my lunch get cold, more soon :-) $\endgroup$ – Jyrki Lahtonen Dec 28 '14 at 12:31
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    $\begingroup$ What you say, Jarki, is clear and correct, I'm just still not sure in which step above is the mistake :-) Probably in the fourth $=$? $\endgroup$ – Peter Franek Dec 28 '14 at 12:33
  • $\begingroup$ @JyrkiLahtonen Thank you Jyrki, but why isn't it true that applying $\varphi_\pi$ to an $n$-tuple results in the $n$-tuple obtained by applying $\pi^{-1}$ to all subscripts? $\endgroup$ – user1337 Dec 28 '14 at 12:35
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    $\begingroup$ As pointed out by others, it applies to positions, not subscripts. $\endgroup$ – Jyrki Lahtonen Dec 28 '14 at 12:46

It is easier to work backward.

You can think of the $k^\text{th}$ element of $\phi_\pi(\mathbf{g})$ as taking the $\pi^{-1}(k)$-th element of $\mathbf{g}$.

So $\phi_{\pi_1}(\phi_{\pi_2}((g_1,g_2,\ldots,g_n)))=\phi_{\pi_1}((g_{\pi_2^{-1}(1)},g_{\pi_2^{-1}(2)},\ldots,g_{\pi_2^{-1}(n)}))$

And then the $k^\text{th}$ element of the result will be the $\pi_1^{-1}(k)$-th element of that in the brackets, which is $g_{\pi_2^{-1}(\pi_1^{-1}(k))}$.

So $\phi_{\pi_1}((g_{\pi_2^{-1}(1)},g_{\pi_2^{-1}(2)},\ldots,g_{\pi_2^{-1}(n)}))=(g_{\pi_2^{-1}(\pi_1^{-1}(1))},g_{\pi_2^{-1}(\pi_1^{-1}(2))},\ldots,g_{\pi_2^{-1}(\pi_1^{-1}(n))})$.

  • $\begingroup$ Harder to type, yes, but addresses the problem. +1. :-) $\endgroup$ – Jyrki Lahtonen Dec 28 '14 at 12:45

The problem is in the fourth equation. The action of a permutation permutates positions, not subscripts. For an example, let $\pi=(1,2,3)$, then $$ \varphi_\pi (g_2,g_1,g_3)=(g_3,g_2,g_1) $$ and not $$ (g_{\pi^{-1}(2)}, g_{\pi^{-1}(1)}, g_{\pi^{-1}(3)})=(g_1, g_3, g_2). $$

  • $\begingroup$ You beat me to it. +1, of course :-) $\endgroup$ – Jyrki Lahtonen Dec 28 '14 at 12:40
  • $\begingroup$ A very nice illustration, thanks! $\endgroup$ – user1337 Dec 28 '14 at 12:41
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    $\begingroup$ My answer was hard to type, haha. $\endgroup$ – user1537366 Dec 28 '14 at 12:43

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