here I am referring to the Problem 8 of the Exercise for Section 5.1 from Dummit Foote. The problem refers to the previous problem which reads:

Let $ G_{1},\dots,G_{n} $ be groups and let $ \sigma\in S_{n} $ be fixed. Prove that the map $$ \varphi_{\sigma}:G_{1}\times \dots\times G_{n}\to G_{\sigma^{-1}(1)}\times \dots\times G_{\sigma^{-1}(n)} $$ defined by $$ \varphi_{\sigma}(g_1,\dots,g_n)=\big(g_{\sigma^{-1}(1)},\dots,g_{\sigma^{-1}(n)}\big) $$ is an isomorphism.

Now problem 8 reads as the following:

In the above exercise, let $ G_1=\dots=G_{n} $, and call $ G=G_1\times \dots\times G_{n} $. Then show that for every permutation $ \sigma\in S_{n} $, the map $ \varphi_{\sigma} $ is an automorphism on $ G $. Also show that the map $ \sigma\mapsto\varphi_{\sigma} $ is an injective homomorphism of $ S_{n} $ into $ \mathscr{A}(G) $, where $ \mathscr{A}(G) $ is the group of automorphisms of $ G $.

I am having trouble with the homomorphism part. I know that I am making a very silly mistake, but cannot find the same. My argument is the following:

For any arbitrary element $ (g_1,\dots,g_n)\in G $ and for arbitrary permutations $ \sigma,\tau\in S_n $, we have $$\begin{align} \varphi_{\sigma\circ\tau}(g_1,\dots,g_n) &= \big(g_{(\sigma\circ\tau)^{-1}(1)}, \dots,g_{(\sigma\circ\tau)^{-1}(n)}\big)\\ &= \big(g_{(\tau^{-1}\circ\sigma^{-1})(1)},\dots,g_{(\tau^{-1}\circ\sigma^{-1})(n)}\big)\\ &= \big(g_{(\tau^{-1}(\sigma^{-1})(1))},\dots,g_{(\tau^{-1}(\sigma^{-1})(n))}\big)\\ &= \varphi_{\tau}\big(g_{\sigma^{-1}(1)},\dots,g_{\sigma^{-1}(n)}\big)\\ &= \varphi_{\tau}\big(\varphi_{\sigma}(g_1,\dots,g_n)\big)\\ &= (\varphi_{\tau}\circ\varphi_{\sigma})(g_1,\dots,g_n). \end{align}$$ Therefore, $ \varphi_{\sigma\circ\tau}=\varphi_{\tau}\circ\varphi_{\sigma} $.

So, the order of the maps has reversed. I cannot seem to find a mistake. Please help.


This is tricky, and the computation has to be done carefully. There is a bit of a surprise in it. Here we go:

The $j$th coordinate of $\varphi_{\rho}(g_1,\ldots,g_n)$ is the $\rho^{-1}(j)$th coordinate of $(g_1,\ldots,g_n)$.

So the $j$th coordinate of $\varphi_{\tau}(\varphi_{\sigma}(g_1,\ldots,g_n))$ is the $\tau^{-1}(j)$th coordinate of $\varphi_{\sigma}(g_1,\ldots,g_n)$.

The $\tau^{-1}(j)$th coordinate of $\varphi_{\sigma}(g_1,\ldots,g_n)$ is the $\sigma^{-1}(\tau^{-1}(j))$th coordinate of $(g_1,\ldots,g_n)$.

That is, $$\begin{align*} \varphi_{\tau}\circ\varphi_{\sigma}(g_1,\ldots,g_n) &= \left(g_{\sigma^{-1}(\tau^{-1}(1))}, g_{\sigma^{-1}(\tau^{-1}(2))},\ldots,g_{\sigma^{-1}(\tau^{-1}(n))}\right)\\ &=\left(g_{\sigma^{-1}\circ\tau^{-1}(1)}, g_{\sigma^{-1}\circ\tau^{-1}(2)},\ldots, g_{\sigma^{-1}\circ\tau^{-1}(n)}\right)\\ &= \left( g_{(\tau\circ\sigma)^{-1}(1)},\ldots,g_{(\tau\circ\sigma)^{-1}(n)}\right)\\ &=\varphi_{\tau\circ\sigma}(g_1,\ldots,g_n). \end{align*}$$

The error in your calculation is in line 4.

  • $\begingroup$ How does your first line equality "$\varphi_{\tau}(\varphi_{\sigma}(g_1,\ldots,g_n)) = \left(g_{\sigma^{-1}(\tau^{-1}(1))}, g_{\sigma^{-1}(\tau^{-1}(2))},\ldots,g_{\sigma^{-1}(\tau^{-1}(n))}\right)$" formally follow from the definition $\varphi_{\rho}(g_1,\ldots,g_n):=(g_{\rho^{-1}(1)},\dots,g_{\rho^{-1}(n)})$? By "formally" I mean without any prior explanation, which though useful, shouldn't be necessary. $\endgroup$
    – CAB
    Sep 22 at 11:55
  • $\begingroup$ @CAB: It's not a "prior explanation", it's the definition. It is a formal. The computation is done from the inside out, and you need to understand how $\tau$ acts. The definition isn't "apply $\tau^{-1}$ to the index". The definition is "move the entry in the $\tau^{-1}(j)$ position to the $j$ position". $\endgroup$ Sep 22 at 12:35
  • 1
    $\begingroup$ @CAB: Two tuples are equal if and only if their $j$th projections are equal for all $j$.By definition, $\pi_j\circ\varphi_{\rho}=\pi_{\rho^{-1}(j)}$. So $\pi_j\circ\varphi_{\tau}\circ\varphi_{\sigma}=\pi_{\tau^{-1}(j)}\circ\varphi_{\sigma}=\pi_{\sigma^{-1}(\tau^{-1}(j))}$. Thus, $\pi_j(\varphi_{\tau}\circ\varphi_{\sigma}) = \pi_J(\varphi_{\tau\circ\sigma})$ for all $j$. $\endgroup$ Sep 22 at 12:42
  • $\begingroup$ This latter approach made the point clear to me. Thanks. $\endgroup$
    – CAB
    Sep 22 at 13:15

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