Every infinite group $G$ has a subgroup $H$ that is non-trivial ($H \ne G, \lbrace e \rbrace$).

Proof: This will be a proof by contradiction. So we will assume every subgroup is trivial and bring the cyclic subgroups to the table.

For the non-identity $\forall x \in G$, $\langle x \rangle \ne \lbrace e \rbrace$. Hence $\langle x \rangle = G.$ Thus every non-identity element of $G$ must be a generator of $G$. Then it must be possible to write any element as the exponent of another. For $\forall y \in G$,

$$y =x^n.$$

Since $x^2 \in G$ by closure, it is also a generator and we must be able to write $x$ as an exponent of $x^2$. Yet this is not possible unless our group is finite. So our assumption must be false and our conjecture must be true. $\square$

Is my proof watertight? And is it rigorous enough? Thanks for your reviews.

  • $\begingroup$ You do not need to phrase it as a proof by contradiction. Take $x\in G$, $x\neq e$. If $x^2=e$, then $\langle x\rangle\neq G$, since $\langle x\rangle$ is finite. If $x^2\neq e$, and $x\in\langle x^2\rangle$, then $x=x^{2k}$ for some $k$, so $x^{2k-1}=e$, hence $\langle x\rangle$ is finite, nontrivial, and distinct from $G$. And if $x\notin \langle x^2\rangle$, $x^2\neq e$, then $\langle x^2\rangle$ is a proper subgroup of $G$ and you are done. $\endgroup$ – Arturo Magidin Oct 19 '19 at 1:33

The idea is right, but you should avoid claims like

Yet this is not possible unless our group is finite.

since this doesn't give your reason for the "unless."

In your case, you perhaps could explicitly note that $G = \langle x \rangle$ implies $x$ cannot have finite order. Then show that $x \in \langle x^2\rangle$ implies $x$ has finite order.

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There is a tiny hole: $x^2$ is not a generator of $G$ if it is equal to the identity element $e$. So you also need to consider the separate case that $x^2=e$. Otherwise, it looks good.

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  • 1
    $\begingroup$ $x^2$ can't be equal to the identity element because by assumption $x$ is a generator of an infinite group, so that exponents of $x$ are all distinct (otherwise there is no way resulting cyclic group can be infinite too). Since $x^0 = e$, it follows that $x^2 \ne e$. $\endgroup$ – İbrahim İpek Oct 18 '19 at 21:22

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