This is Exercise 1.5.10 of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.
The Details:
On page 11, ibid., the set product of subsets $X,Y$ of a group is defined as
$$XY=\{xy\mid x\in X, y\in Y\}.$$
On page 17, ibid., for groups $G,H$, the definition of ${\rm Hom}(G, H)$ is given as the set of all homomorphisms from $G$ to $H$.
The definition of a direct product on pages 20 to 21, ibid., is a little more involved.
Let $\{G_\lambda\mid \lambda\in\Lambda\}$ be a given set of groups. The cartesian (or unrestricted direct) product,
$$C=\underset{\lambda\in\Lambda}{{\rm Cr}}\, G_\lambda,$$
is the group whose underlying set is the set product of the $G_\lambda$s [. . .] and whose group operation is multiplication of components: thus
$$(g_\lambda)(h_\lambda)=(g_\lambda h_\lambda),$$
$g_\lambda, h_\lambda\in G_\lambda$. [. . .]
The subset of all $(g_\lambda)$ such that $g_\lambda=1_\lambda$ for almost all $\lambda$ [. . .] is called the external direct product,
$$D=\underset{\lambda\in\Lambda}{{\rm Dr}}\, G_\lambda,$$
[. . .] In case $\Lambda=\{\lambda_1,\dots,\lambda_n\}$, a finite set, we write
$$D=G_{\lambda_1}\times\dots\times G_{\lambda_n}.$$
Of course $C=D$ in this case.
On page 26, ibid., we have that
The set of all automorphisms of [a group] $G$ is denoted by
$${\rm Aut}(G).$$
It is a group under function composition.
The Question:
This is on page 30, ibid.
Let $G=G_1\times\dots\times G_n$ where the $G_i$ are abelian groups. Prove that ${\rm Aut}(G)$ is isomorphic with the group of all invertible $n\times n$ matrices whose $(i,j)$ entries belong to ${\rm Hom}(G_i, G_j)$, the usual matrix product being the group operation.
Thoughts:
I'm used to thinking of the direct product of groups as the group defined by a certain presentation, or else simply as the obvious generalisation of $$H\times K=\{(h,k)\mid h\in H, k\in K\}$$ under the operation $(h,k)\cdot(h',k')=(hh', kk')$.
My linear algebra is a little rusty, which is probably why I'm struggling here.
Examples:
When $n=1$:
We have just the abelian group $G$. The automorphism group is
$${\rm Aut}(G)=\{\varphi\in {\rm Hom}(G,G)\mid \varphi \text{ is an isomorphism}\}$$
and the $1\times 1$ matrices are simply $(\varphi)$ with $(\varphi)^{-1}=(\varphi^{-1})$.
The isomorphism is trivial.
When $n=2$:
We have $G=G_1\times G_2$. Now the $(i,j)$ entries of the $2\times 2$ matrices are such that . . .
Wait a Minute . . .
What operation is the addition of the homomorphisms?
I guess this is my main stumbling block.
Let's try working it out & see . . .
So, we have
$${\rm Aut}(G)\stackrel{?}{\cong}\left\{
\begin{pmatrix}
\varphi_{(1,1)} & \varphi_{(1,2)}\\
\varphi_{(2,1)} & \varphi_{(2,2)}
\end{pmatrix} : \varphi_{(1,1)}\stackrel{?}{\times}\varphi_{(2,2)}\stackrel{?}{-} \varphi_{(1,2)}\stackrel{?}{\times}\varphi_{(2,1)}\neq \stackrel{?}{0} \right\}.$$
Okay, so I'm not sure on multiplication either, nor am I sure of what the zero element is.
This is a mess. I'm sorry.
Please help 🙂
Best Answer
Hints:
When $G$ is any group and $B$ is an abelian group, then $\mathrm{Hom}(G,B)$ is more than just a set, it is actually an abelian group. The operation is "pointwise addition", so that $(f+g)(a) = f(a)+g(a)$ for all $a\in G$. This is a homomorphism as well, since $$\begin{align*} (f+g)(a+a') &= f(a+a') + g(a+a')\\ &= f(a)+f(a') + g(a)+g(a')\\ &= f(a)+g(a)+f(a')+g(a')\\ &= (f+g)(a)+(f+g)(a'). \end{align*}$$ The identity element of this abelian group is the zero map, sending everything to $0_B$, and the additive inverse of $f$ is the function $g(a)=-f(a)$ for all $a\in A$. (When $G=B$, this is actually a ring, with the multiplication being composition of functions).
For any family of groups $\{G_i\}_{i\in I}$, a morphism $f$ from a group $H$ into the direct product $\prod_{i\in I}G_i$ is equivalent to a family of morphisms $\{f_i\}_{i\in I}$, with $f_i\colon H\to G_i$. This is the universal property of the product: given the family $\{f_i\}_{i\in I}$, there exists a unique morphism $F\colon H\to \prod G_i$ such that $f_i =\pi_i\circ F$, where $\pi_i$ is the projection onto the $i$th coordinate. Namely, $F(h) = (f_1(h),\ldots,f_n(h))$.
For a finite family of abelian groups, we also have maps going the other way. The embeddings $\iota_i\colon A_i\to \prod_{j\in J}A_j$ that sends $a\in A_i$ to the element of the product that has $a$ in the $i$th coordinate and $0$s in all the other coordinates have the following universal property: given any abelian group $K$, and given any morphisms $g_i\colon A_i\to K$ for each $i$, there exists a unique morphism $G\colon \prod A_i\to K$ such that $g_i = G\circ \iota_i$ for each $i$. Namely, $G(a_1,\ldots,a_n) = g_1(a_1)+\cdots + g_n(a_n)$. Thus, maps from a finite direct product of abelian groups are equivalent to a family of maps from the factors.
Putting those together, it should be clear that any homomorphism $G_1\times\cdots\times G_n\to G_1\times\cdots\times G_n$ corresponds to a family of morphisms $\varphi_{ij}\colon G_j\to G_i$. You should then verify that if you arrange them in a matrix in the "obvious" way, and what the corresponding morphism is, then the image of an element $(g_1,\ldots,g_n)$ under the corresponding morphism $\varphi\colon G_1\times\cdots\times G_n\to G_1\times\cdots\times G_n$ is given by computing $$\left(\begin{array}{ccc} \varphi_{11}& \cdots & \varphi_{1n}\\ \vdots & \ddots & \vdots \\ \varphi_{n1} & \cdots & \varphi_{nn} \end{array}\right) \left(\begin{array}{c} g_1\\ \vdots\\ g_n\end{array}\right).$$
That will give you a description of the homomorphisms. It also tells you that the composition of homomorphisms "should" correspond to matrix multiplication. And that will tell you exactly when a homomorphism is actually an automorphism.