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I was wandering if anyone could tell me how my proof looks and it makes since. Here is the following question: Consider the infinite direct product $H = \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times$ $...$. I want to show that it is not finitely generated.

Proof: Suppose $H$ is generated by finitely many elements $k$. Since each element in $H$ has order $2$, the group generated by these elements has order of at most $2^k$. This is a contradiction since $H$ has infinite order, and such a set could not generate $H$. Therefore, $H$ is not finitely generated.

Mick Smith
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  • It's unnecessary to phrase it as a proof by contradiction. You are just proving, directly, that every finitely generated subgroup is finite, so $H$ (being infinite) is not finitely generated. – Qiaochu Yuan Sep 27 '20 at 23:56

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Your proof seems that nothing wrong, but one point is missing. You have to mention that $H$ is abelian when you deduce that generated group, say $\langle g_1, \cdots, g_k \rangle$, is of order at most $2^k$. There is no guarantee that $g_ig_j$ is of finite order in general.

Orat
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  • Suppose $H$ is generated by finitely many elements $k$. Since each element in $H$ has order $2$, the group $<g_{1}, ... , g_{k}>$ generated by these elements has order of at most $2^k$. This is a contradiction since $H$ has infinite order, and such a set could not generate $H$. Therefore, $H$ is not finitely generated. How does this proof sound now? – Mick Smith Sep 05 '13 at 04:16
  • Is something differ from original one? Think about dihedral group $D_n = \langle r, s \mid r^n = s^2 = (rs)^2 = 1 \rangle$ for example (though this may be not good enough example). This group also generated by involutions (elements of order 2) $rs, s$; and the order of generated group is $2n$. So $2n > 2^2$ if $n \geq 3$. This is the case I mentioned in the answer by and large. So you have to mention that $H$ is abelian in the proof. – Orat Sep 05 '13 at 11:45
  • If I mention that $H$ is abelian in the proof, will the proof be good then? – Mick Smith Sep 05 '13 at 17:27
  • @MickSmith Absolutely. Though I prefer to show "Let $g_1, \cdots, g_k \in H$ and $K := \langle g_1, \cdots, g_k \rangle$ ... $K$ has a finite order at most $2^k$, which implies $K \neq H$" to avoid proof by contradiction. – Orat Sep 05 '13 at 20:47