2
$\begingroup$

Suppose that $G_1$, $G_2$, $H_1$ and $H_2$ are non-trivial groups. Under what conditions do we have $$G_1 \ast H_1 \cong G_2 \ast H_2?$$

Is it the case that if $G_1 \cong G_2$ then we only have such an isomorphism if $H_1 \cong H_2$?

This may be quite a wide and complicated question, so if a full answer is difficult, I'm most interested in the following two cases:

1) $G_1 \cong G_2$ and $H_1$, $H_2$ are both finite.

2) $G_1 \cong G_2$ and $H_1$, $H_2$ are both free.

  • 0
    You might want to restrict attention to finitely generated groups, since otherwise it's easy to construct examples in which $H_1$ and $H_2$ are not isomorphic. Let $G_1$ be a free product of infinitely many copies of $H_1$ and of $H_2$. – 2017-01-01
  • 0
    Ah, of course. So for example, if $G_1 \cong G_2 \cong F_{\mathbb{N}}$ and $H_1 \cong F_3$, $H_2 \cong F_5$, then $G_1 \ast H_1 \cong G_2 \ast H_2 \cong F_{\mathbb{N}}$. – 2017-01-01

1 Answers 1

2

The magic words are Grushko decomposition Theorem.

  • 0
    @user60589 The link provides a complete answer to your question. If you don't want it, suit yourself. – 2017-01-03