# Every infinite group has a non-trivial subgroup

Theorem:

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.

• 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. – Arturo Magidin Oct 19 '19 at 1:33

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.
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.
• $x^2$ 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 $x^2 \ne e$. – İbrahim İpek Oct 18 '19 at 21:22