# Is there a simple abelian group $G$ with infinite order?

I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.

There is the following problem in this book:

On p.80
Problem 8:
Show that a simple abelian group $$G \neq \{e\}$$ is a cyclic group whose order is prime.

Did the author intend the following problem?

If this problem were the following, I could solve it:

Problem 8':
Show that a simple finite abelian group $$G \neq \{e\}$$ is a cyclic group whose order is prime.

Proof:
Because $$G \neq \{e\}$$, there is an element $$g \in G$$ which is not equal to $$e$$.
$$H := \{g^i | i \in \mathbb{Z}\}$$ is a subgroup of $$G$$ and $$G$$ is abelian.
So $$H$$ is a normal subgroup of $$G$$.
And $$G$$ is simple.
And $$H \ni g \ne e$$.
So $$H = G$$.
Let $$n := \#H = \#G$$.
Then, $$n$$ is prime.
If $$n$$ is not prime, then we can write $$n = d d'$$, $$1 < d < n$$, $$1 < d' < n$$.
Then $$H' := \{(g^d)^i | i \in \mathbb{Z}\}$$ is a subgroup of $$G$$ whose order is $$d'$$.
So, $$H'$$ is neither $$\{e\}$$ nor $$G$$.
Becasue $$G$$ is abelian, $$H'$$ is a normal subgroup of $$G$$.
And $$G$$ is simple.

• No, the problem as stated in the book, is correct. You are supposed to prove that any simple abelian group is finite. It is not difficult, so you should try and do it yourself. Feb 1, 2019 at 9:20
• Thank you very very much, for your answer. Derek Holt. I will try. Feb 1, 2019 at 9:21

Suppose that $$G$$ is simple Abelian, if $$x$$ in $$G$$ has an infinite order the group generate by $$2x$$ is a proper normal subgroup since it does not contain $$x$$.

If $$x$$ has a finite order, there exists $$y$$ not in $$H$$ the group generated by $$x$$, and $$H$$ is normal and proper.

• Thank you very much, Tsemo Aristide. But I cannot understand your answer. Suppose $G$ is simple infinite Abelian and $x \in G-\{0\}$. Then, $G = \{i x | i \in \mathbb{Z}\}$ because $G$ is simple. So $x$ has an infinite order. $\{i 2 x | i \in \mathbb{Z}\}$ is a proper normal subgroup of $G$ since it does not contain $x$. But $G$ is simple. This is a contradiction. Feb 1, 2019 at 10:26
• Sorry, I understand your answer now. Feb 1, 2019 at 10:32
• Suppose that $G$ is simple infinite Abelian. If $x \in G$, $x$ has an infinite order or a finite order. If $x$ has an infinite order, then the group generated by $2x$ is a proper normal subgroup since it does not contain $x$. This is a contradiction. If $x$ has a finite order, then there exists $y$ not in the group generated by $x$, because $G$ is infinite. So $H$ is normal and proper. This is a contradiction. Thank you very very much, Tsemo Aristide. Feb 1, 2019 at 10:37

Since $$G$$ is abelian, all the subgroups are normal. So $$G$$ simple abelian means there are no proper subgroups.

So take $$x\neq e \in G$$.

Claim: $$\langle x\rangle =G$$.

If not, $$\langle x\rangle$$ is a proper normal subgroup. Contradiction.

So $$G$$ is cyclic, and contains no proper subgroups.

Claim: $$\mid G\mid= p$$, where $$p$$ is prime.

Suppose not. Then either $$\mid G\mid=n$$, where $$n=rs$$ for some $$r,s\neq1$$, or $$G$$ is infinite. Take $$x\neq e$$. In the first case, $$\langle x^r\rangle \le G$$ is proper. Contradiction. In the second case, $$\langle x^2\rangle\le G$$ is proper (this requires a little argument which I will leave to you). Also a contradiction.

• Thank you very much, for your answer, Chris Custer. Feb 1, 2019 at 11:28