2
$\begingroup$

Prove that $A_{5}$ has a subgroup of order $12$.

What I have so far: The order of an alternating subgroup $A_{n}$ = $n!/2$. So the order of $A_{5}$ is $60$. We know that the order of a subgroup must be divide the order of the group, and this is the case since $12$ divides $60$.

Also, I have another question. Are all alternating groups cyclic?

$\endgroup$

2 Answers 2

Reset to default
4
$\begingroup$

Consider the Alternating group $A_{4}$, it has order 12, and it's in $A_{5}$. Aletnaring group $A_{n}$ is not cyclic for $n >3$

$\endgroup$
2
  • $\begingroup$ Why is $A_{4}$ a subgroup of $A_{5}$? Is it always the case that a smaller alternating group will be a subgroup of a larger alternating group? $\endgroup$
    – EmaLee
    Apr 13, 2015 at 0:16
  • $\begingroup$ Never mind, someone else answered this question. $\endgroup$
    – EmaLee
    Apr 13, 2015 at 0:17
3
$\begingroup$

Are alternating groups cyclic? Well, all cyclic groups are abelian (you should prove this). Is the alternating group abelian?

As a hint to your subgroup of order $12$ question, note that $A_n \subset A_m$ for all $n < m$.

$\endgroup$

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.