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Given $G$, $H$, $G'$, and $H'$ are cyclic groups of orders $m$, $n$, $m'$, and $n'$ respectively.

If $G*H$ is isomorphic to $G'* H'$, I would like to show that either $m = m'$ and $n = n'$ or else $m = n'$ and $n = m'$ holds. Where * denotes the free product.

My approach:

$G*H$ has an element of order $n$, thus $G' * H'$ has one too.

But already the next step is not clear to me, should I show that there is an element of length $> 1$ which has infinite order or what would be the right approach here?

Thank you.

  • 1
    In group theory, $\ast$ means [the free product](http://en.wikipedia.org/wiki/Free_product). Did you mean that, or did you mean $\times$, [the direct product](http://en.wikipedia.org/wiki/Direct_product_of_groups)?2011-12-15
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    Do you know that all elements of finite order are conjugate to an element of either $G$ or $H$? That should help show the first part. Now also prove that the if $K$ is the normal closure of $H$, then $(G*H)/K\cong G$.2011-12-15
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    yes I ment the free product2011-12-15

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