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Let $C_n$ be the cyclic group of order $n$. We know that $C_4$ is an extension of $C_2$ by $C_2$, so there must be a way of combining $C_2$ with itself in order to obtain $C_4$. But the only way I know of combining a group with another group is by way of the product operations, e.g. the direct, semi-direct, and free products. Yet, both the direct and the semi-direct product of $C_2$ by $C_2$ yield the Klein four group $V_4$, and if I'm not mistaken the free product will also not yield $C_4$ (though I'm less sure about this). Am I missing something?

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    $\begingroup$ Concerning the use of free products, $C_2 \ast C_2 = D_\infty$, so I'd guess your intuition is right. $\endgroup$ – theyaoster Apr 17 '17 at 19:38
  • $\begingroup$ @BrianYao Thanks, that confirms my suspicion. $\endgroup$ – Nagase Apr 17 '17 at 19:40
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    $\begingroup$ What does 'extension' mean here? Do we mean that $C_4$ has $C_2$ as a subgroup, and that has index (?) $2$? In that case, it seems more like we want the relation $x^2 = e$ for some $x$, and $y^2 = x$ for some $y$. These two relations present $C_4$ iirc. $\endgroup$ – Artimis Fowl Apr 17 '17 at 22:11
  • $\begingroup$ @ArtimisFowl - $G$ is an extension of $H$ by $K$ if there is a group $K' \leq G$ such that $K'$ is normal in $G$, $K' \simeq K$, and $G/K' \simeq H$. As Rotman says in his introduction to group theory, p. 155, this means that $G$ is some kind of "product" or combination of $H$ and $K$. $\endgroup$ – Nagase Apr 17 '17 at 23:44
  • $\begingroup$ Is this not an inner semidirect product? $\endgroup$ – Santana Afton Apr 18 '17 at 1:47

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