# Homomorphism between $S_n$ and the group of Automorphisms of a group of direct product

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.

## 1 Answer

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.

• 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.
– CAB
Sep 22 at 11:55
• @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". Sep 22 at 12:35
• @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 $j$. Sep 22 at 12:42
• This latter approach made the point clear to me. Thanks.
– CAB
Sep 22 at 13:15