Proof that all abelian simple groups are cyclic groups of prime order

Question

Just wanted some feedback to ensure I did not make any mistakes with this proof. Thanks!

Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| > 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has either infinite order or even order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has odd order, and by Feit-Thompson, $G\cong Z_p$ for some prime $p$.

Edit: Thanks, I see that Feit-Thompson is too much.

Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| > 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has infinite order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has finite order. Suppose $x$ has composite order $n$ so for some $p > 1$ that divides $n$, $\langle x^p \rangle$ is a proper non-trivial subgroup of $G$, so $G$ is not simple. So $G$ is a cyclic group of prime order.

Answer 1

Er okay: suppose $G$ is simple, abelian. Then any $g \in G$ has to generate the whole thing, i.e. we have $\mathbb{Z} \twoheadrightarrow G$. $\mathbb{Z}$ is not simple (consider say $2\mathbb{Z}$), so $G \simeq \mathbb{Z}/n \mathbb{Z}$ for some $n$. If $n$ is not prime, take any $p \mid n$, we have $\langle p\rangle \subseteq \mathbb{Z}/n\mathbb{Z}$ is a proper subgroup.

Answer 2

Definitely you're swatting a fly with a nuclear weapon. The Feit–Thompson theorem is not easy to prove, to put it mildly. But it's pretty easy to prove that all abelian simple groups are cyclic groups of prime order.

Suppose it's not cyclic. It's not generated by any one element. Consider $a\ne\mathrm{identity}$. Then some $b$ is not in the subgroup generated by $a$. Think about the subgroups generated by these two elements and see if you can find a nontrivial normal subgroup.

Now suppose the order is $n=k\ell$ and $k,\ell>1$. Let $a\ne\mathrm{identity}$. Think about the group generated by $a^k$ and see if you can find a nontrivial normal subgroup.

And the case of infinite order is not hard. ${{}}$

Answer 3

Every subgroup of an abelian group is normal.Thus, a simple abelian group must have no proper nontrivial subgroup. We also note that no proper nontrivial subgroup implies cyclic of prime order.

Answer 4

Let $G$ be an abelian simple group with $|G| > 1$. We want to prove that $G$ is a cyclic group of prime order.

First note that because $G$ is abelian, then every subgroup of $G$ is normal.

Let $x \in G$ and $x \neq 1$, so $\langle x \rangle \leq G$. But because $G$ is simple, $\langle x \rangle$ must be the $G$ itself as there couldn't be any other normal subgroup of $G$, or simply $\langle x \rangle = G$.

If $|G|$ is not finite, then we could have e.g. $\langle x^2 \rangle \neq \langle x \rangle$ as another proper normal subgroup of $G$ which is a contradiction becase G is simple. (It's a proper normal subgroup because $|x^2| < |x|$.)

Hence $G = \langle x \rangle$ must be a finite group.

Let $p = |G|$. Assume $p = mn$ where $m > 1$ and $n > 1$. If that's the case, then we could have $\langle x^m \rangle$ or $\langle x^n \rangle$ as another proper normal subgroup of $G$, which is again a contradiction.

Thus $G = \langle x \rangle$ and $|G| = |x| = p$ where $p$ is a prime number.

Answer 5

I found a simple enough solution using the classification theorem for finitely generated abelian groups and some ideas I saw earlier on in these responses. First note that any subgroup in an abelian group is a normal subgroup, and hence if an abelian group $G$ is simple, then we must prove it has no subgroups. If $G$ is a finitely generated abelian group, then $G \cong \mathbb{Z}_{m_1}\times \cdots \mathbb{Z}_{m_k}\times \mathbb{Z}^s$, where $m_i\mid m_{i+1}$ for $i=1,\cdots k-1$, $\mathbb{Z}^s={1}$ for $s=0$, $\mathbb{Z}^s=\mathbb{Z}\times\cdots\times\mathbb{Z}$ $s$ times. Clearly if $G$ is infinite, then $\exists\langle2\rangle\in\mathbb{Z}$ where $\langle2\rangle\lneq\mathbb{Z}$, hence is not simple. If $G$ is finite, clearly $k=1$. If $k>1$, there exists an element $a\in G$ such that the order of $a\mid |G|$ and is less than $|G|$, hence $\langle a \rangle\lneq G$ so $G$ is not simple. So $G\cong\mathbb{Z}_m$. If $m$ is composite, say $m=xy$, then there exists an element $a\in G$ such that the order of $a$ is $x$, and hence $\langle a\rangle\lneq G$. So $m$ must be prime. Hence, all simple abelian groups $G\cong \mathbb{Z}_p$ for some prime $p$.