This is part of Exercise 3.3.12 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0 and this search, it is new to MSE. I am not familiar with representation theory nor is that topic covered in the book so far.
The Details:
Since definitions vary, on page 15, ibid., paraphrased, it states that
A subgroup $N$ of $G$ is normal in $G$ if one of the following equivalent statements is satisfied:
(i) $xN=Nx$ for all $x\in G$.
(ii) $x^{-1}Nx=N$ for all $x\in G$.
(iii) $x^{-1}nx\in N$ for all $x\in G, n\in N$.
We call a group simple if $G\neq 1$ and $1$ and $G$ are the only normal subgroups of $G$. This definition is equivalent to the one found on page 16.
Robinson defines completely reducible in terms of $\Omega$-groups on page 85; however, for $\Omega=\varnothing$, the definition is equivalent to saying
a group is completely reducible if it is a direct product of a possibly infinite family of simple groups.
The Question:
Prove that the image of a completely reducible group is completely reducible.
I assume here that "image" refers to "homomorphic image".
Thoughts:
I start out by supposing, for a group $G$, that
$$G=\varphi \left(\underset{i\in\mathcal{I}}{{\rm Dr}}\, G_i\right),\tag{1}$$
where $\varphi$ is a group homomorphism, each $G_j$ is a simple group, and $\mathcal{I}$ is a (potentially infinite) index set. Let $H=\underset{i\in\mathcal{I}}{{\rm Dr}}\, G_i$.
I suspect, from experience and $(1)$, that I should aim for
$$G\cong\underset{i\in\mathcal{I}}{{\rm Dr}}\, \left(\varphi\rvert_{G_i} (G_i)\right),\tag{2}$$
where $\varphi\rvert_{G_j}$ the restriction of $\varphi$ to $G_j$.
I suppose there are two stages:
Go from $(1)$ to $(2)$. I think this follows from the fact that both $\varphi$ and each $\varphi\rvert_{G_j}$ are homomorphims.
Show that each $\varphi\rvert_{G_j} (G_j)$ is simple. I think that follows from the first homomorphism theorem, since the kernel of $\varphi\rvert_{G_j}$ is normal in $G_j$.
I think I could solve this on own if I could follow steps like this. My confidence is low, however, because Robinson defines the direct product on pages 20 to 21, ibid., like so:
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.
I'm not used to it.
Moreover, I want to move on to the rest of the book. I've given this problem a couple of days and I reckon that's enough for half an exercise.
The kind of answer I'm looking for is a full solution or, if not, a list of strong hints.
Please help :)