If $G_1,...,G_n$ are groups and $G=\prod_{i=1}^nG_i$, is it true that each $G_i$ is isomorphic to some subgroup of $G$?

Question

If $G_1,...,G_n$ are groups and $G=\prod_{i=1}^nG_i$, is it true that each $G_i$ is isomorphic to some subgroup of $G$? How do you prove this? Inductively? Is there a non-inductive method?

Answer 1

It is indeed, and this works for infinite products as well as finite ones. Suppose $\displaystyle G = \prod_k G_k$ and consider the set $H = \{(e_{G_1}, e_{G_2}, ..., g_j, ..., e_{G_n}, ...) \ | \ g_j \in G_j\}$ where $e_{G_k}$ denotes the identity in $G_k$. This is to say, consider the set of elements of $G$ where each coordinate is the identity in its respective group except for the $j^{\text{th}}$ coordinate, which can be any element of $G_j$. You can prove that $H$ is a subgroup of $G$, and there's a fairly natural isomorphism between $H$ and $G_j$; I'll leave the details to you.

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Now, if you really wanted to, you can also proceed with strong induction to get the result for finite products. The base case is trivial: any group $G_1$ is a subgroup of itself. Then suppose this holds for $\displaystyle \prod_{k=1}^n G_k$ for all $1 \leq k \leq n$. For the induction step, we consider $\displaystyle G = \prod_{k=1}^{n+1} G_k$. If you want to show $G_j$ is a subgroup of $G$ for some $1 \leq j \leq n\!+\!1$, first recognize the isomorphism $\displaystyle G \cong \left( \prod_{k \neq j} G_k \right) \times G_j$ (rearranging the order of the product doesn't change the isomorphism class to which $G$ belongs; this is a generalization of the fact that $A \times B \cong B \times A$). Given that we have now written $G$ as a product of two groups, use the induction hypothesis to finish up.

Answer 2

Let $G$ be the product of groups $G_i$, $i\in I$, which by definition means

There exist homomorphisms $\pi_i\colon G\to G_i$ such that for every group $X$ and family of homomorphisms $f_i\colon X\to G_i$, there exists one and only one homomorphism $h\colon X\to G$ such that $f_i=\pi_i\circ h$ for all $i$.

Note that a priori this gives us only $G\to G_i$, not the desired $G_i\to G$. But this universal property gives us a recipe, how to obtain a homomorphism $G_j\to G$: We let $X=G_j$, find a nice family of homomorphisms $f_i\colon G_j\to G_i$ and then obtain our $h\colon G_j\to G$.

What $f_i$ can we pick? For $i\ne j$, the only apparent choice is the zero homomorphism. And for $i=j$ the only apparent non-trivial choice is the identity. So let's indeed pick $f_i=0$ for $i\ne j$ and $f_j=\operatorname{id}_{G_j}$. Then by definition, the resulting homomorphism $h\colon G_j\to G$ has the property $\pi_j\circ h=f_j=\operatorname{id}_{G_j}$. In particular, $h$ is injective and allows us to view $G_j$ as subgroup of $G$.