I'm asking myself what the meaning of a statement like the following is:
Let $G$ be a group and $T_i \subseteq G$ be a family of subgroups of $G$ indexed by a possibly infinite set $I$. Now let $H$ be another group generated by $T_i, i\in I$.
As I understand the above is, that there is a surjective group homorphism
$$\prod_{i \in I} T_i \twoheadrightarrow H,$$
because every element in $H$ comes from some product of elements in $T_i$.
But consider the following case where $H'$ is another group: Let $\varphi$ be an injective group homomorphism
$$\varphi: H' \hookrightarrow \prod_{i\in I} T_i$$
My first question now is:
Is it correct to say that $H'$ is generated by the $T_i$?
Secondly, is in which case can we talk about $\prod_{i \in I} T_i$ being the free product over $I$?
Thank you, Tom
As @H.B. said, $H$ is the smallest subgroup of $G$ that contains each of the $T_i$. There is not necessarily a surjective homomorphism $\phi: \prod T_i \to H$. If you replace $\prod T_i$ with $\oplus T_i$, and you add the assumption that $t_i t_j = t_j t_i$ for all $t_i \in T_i, t_j \in T_j, i \ne j$, then there is a canonical surjective homomorphism $\phi$ (if $|I| = 2$, this is given by $(t_1, t_2) \mapsto t_1 t_2)$. However, without this commutativity condition, $\phi$ will definitely not be a homomorphism, and might not be surjective as a set map.
As examples, consider $G = S_3$. In one example, $T_1$ is generated by $(1 2)$, and $T_2$ is generated by $(1 2 3)$. In the second example, $T_1$ is generated by $(1 2)$ and $T_2$ is generated by $(1 3)$. In both cases, $H = G$. In the first example, $\phi$ is surjective but not a homomorphism. In the second example, $\phi$ isn't even surjective, and it had no hope of being surjective since $ 6 = |H| > |T_1 \times T_2| = 4$.