Does there exist a surjective group homomorphism from G to H such that G doesn't contain a subgroup isomorphic to H?

Question

Earlier this week, my Professor was doing examples of surjective group homomorphisms between different groups $G$ and $H$. In each example, I began to notice a pattern -- in each case, $G$ always contained a subgroup that was isomorphic to $H$.

My question is, is there ever a situation where this is not the case? That is, are there any surjective group homomorphisms between two groups $G$ and $H$ such that $G$ does not contain a subgroup isomorphic to $H$?

Answer 1

Sure, and the easiest example is $\mathbb Z\to\mathbb Z_n=\mathbb Z/n\mathbb Z$ for $n>1$. $\mathbb Z$ contains no nonidentity elements of finite order so it can't have a subgroup isomorphic to the image.

Answer 2

Yes. Take$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&S^1\\&x&\mapsto&(\cos x,\sin x).\end{array}$$The group $(\mathbb R,+)$ has no subgroup isomorphic to $S^1$.

Answer 3

The smallest $G$ for which this is possible is $G\cong Q_8$, the quaternion group.

It has a quotient isomorphic to $C_2\times C_2$ but the only subgroups of $G$ of order $4$ are cyclic.