4
$\begingroup$

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

Assume that $a$ has infinite order in $G$. Then let $y\neq e$ belong to the intersection of all subgroups of $G$ different from $\{e\}$. Then $y=a^n$ and $y$ will also belong to subgroup generted by $a^2$ hence $y=a^{2m}$ for some integers $m$. This contradicts that $a$ has infinte order.

Is this correct?

$\endgroup$
  • $\begingroup$ How do you find that $y=a^n=a^{2m}$? $\endgroup$ – YTS Jan 8 '16 at 15:56
  • $\begingroup$ $y$ will also belong to subgroup generted by $a^2$ $\endgroup$ – Eklavya Jan 8 '16 at 16:01
2
$\begingroup$

To express your idea differently:

If $G$ has an element $a$ of infinite order, then the intersection of all nontrivial subgroups is trivial.

Indeed, this is true for $\mathbb Z$, which is isomorphic to $\langle a \rangle$.

Therefore, if the intersection of all nontrivial subgroups is nontrivial, then $G$ cannot have an element of infinite order and so all elements must have finite order.

$\endgroup$
1
$\begingroup$

Not completely true, but it is the idea precise that $y$ is in the group generated by $a$ so $y=a^n$. $y$ is also in the group generated by $a^{2n}$ so $y={a^{2n}}^m=a^{2nm}$, so $a^{2nm-n}=1$ contradiction

$\endgroup$
  • $\begingroup$ because it is in the intersection of all subgroup of $G$ and the subgroup generated by $a^{2n}$ is a subgroup of $G$. $\endgroup$ – Tsemo Aristide Jan 8 '16 at 16:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.