# Showing that if the intersection of all subgroups other than $\langle e \rangle$ is not $\langle e \rangle$, then every element is of finite order

This is problem 2 on page 46 of I. N. Herstein's Topics in Algebra.

Let $$G$$ be a group such that the intersection of all its subgroups which are different from $$\langle e\rangle$$ is a subgroup different from $$\langle e\rangle$$. Prove that every element in $$G$$ has finite order.

• related Commented Aug 6, 2014 at 16:16
• In case MSE doesn't notify you of edits, I've updated my answer. Commented Aug 6, 2014 at 17:34

Hint: Assume that not every element has finite order; i.e., that there is an element $a$ of infinite order. What can you say about the intersection of the non-trivial subgroups of the cyclic subgroup $\langle a\rangle$.

A simpler answer than my previous one:

Consider

$$H=\bigcap_{g\in G} \langle g\rangle$$

The intersection of cyclic groups is cyclic, so either

(1) $H=\langle h\rangle$ with $\text{ord}_G(h)=n<\infty$

(2) $H=\langle h\rangle\cong \Bbb Z$ whence $\langle h^2\rangle$ is a proper subgroup, contradicting that $H$ was the intersection of all cyclic subgroups of $G$.

So the intersection has finite order. Since $n\ne 1$ by assumption, every $g\in G$ has that $\langle h\rangle \le \langle g\rangle$, but every non-identity element of an infinite cyclic group has infinite order, and clearly $h\in\langle g\rangle$ does not, so $|\langle g\rangle|<\infty$

Let $$a \neq e$$ be an element from that intersection. If $$a^2=e$$, then $$a$$ has order $$2$$. Otherwise, $$a \in \langle a^2 \rangle$$, so $$a=a^{2k}$$, for some integer $$k$$, which implies $$a^{2k-1}=e$$, so $$a$$ has finite order. Either way $$a$$ has finite order.

Now let $$g \in G$$ be arbitrary. Since $$a \in \langle g \rangle$$, it follows that $$g^{\ell}=a$$ for some integer $$\ell$$, so $$g^{\ell \cdot \mathrm{ord}(a)}=e$$, so the order of $$g$$ is finite.

• The expression " if $a$ is not of order $2$, then $a\in \langle a^2 \rangle$" is not correct, perhaps you mean something else? Commented Aug 18, 2023 at 0:41
• @mesel According to the hypothesis, $a$ is contained in any nontrivial subgroup of $G.$ If $\mathrm{ord}(a) \neq 2,$ then $\langle a^2 \rangle$ is a nontrivial subgroup of $G$, and so $a \in \langle a^2 \rangle.$ Commented Aug 18, 2023 at 10:39
• Now, I see what you mean. Commented Aug 18, 2023 at 11:08