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I'd like to prove the following theorem :

Let $G$, $H$, $G'$, $H'$ four groups. If they are finite then $$ \left\lbrace \begin{array}{ll} G\times H \simeq G'\times H' \\ G\simeq G' \end{array}\right. \quad \Rightarrow\quad H\simeq H'.$$

This not stands with infinite groups. Consider for example $G=\mathbb{Z}[X]$, $G'=\bigoplus_{n\geq 0}\mathbb{Z}$ and $H=\mathbb{Z}$, $H'=\lbrace \mathrm{Id} \rbrace$.

Our teacher proved the theorem using the following method : we first show that for all finite groupe $L$ the number of morphism beetwen $L$ and $H$ is the same that beetween $L$ and $H'$. Then we proved that the number of one to one morphism beetwen $H$ and $H$ is the same that beetwen $H$ and $H'$. We concluded by cardinality.

I'm okay with this proof but I find it a bit artificial. I've tried to find this theorem in books but after hours of rechearch I wasn't able to find any book which prove this theorem.

If you see another proof, or if you know where I can find it, please let me know. Thank you for your help.

Ben Grossmann
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Zanzi
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    This: http://www.jstor.org/stable/2317133?seq=1#page_scan_tab_contents and this: http://math.stackexchange.com/q/1485723/133781 might help you. – Xam Apr 10 '17 at 20:15
  • There is a proof I really like of this statement you can find there: http://math.stackexchange.com/a/427640/36434 – Seirios Apr 12 '17 at 07:33
  • Yes @Xam it is exactly what I was looking for. – Zanzi Apr 13 '17 at 14:55

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