Is there a 20-order abelian subgroup of $S_5$?

Question

The title says it all: Is there an abelian subgroup of order 20 of $S_5$, the group of permutations of five elements?

Thanks for reading!

Answer 1

The answer is no.

Note that such a group must have an element $x$ of order $2$ and another element $y$ of order $5$. Since the subgroup is commutative, you can show that $xy$ must have order $10$. But not $\sigma \in S_5$ has order $10$....

P.S. If Cauchy Theorem is too strong to use, it is easy to show that every group $H$ of even order contains an element of order $2$.

Assume by contradiction that $H$ doesn't have an element of order $2$, then $e \in H$, and all the other elements can be paired in pairs of two distinct elements $(x, x^{-1})$. Therefore $H$ has an odd number of elements.

To prove that there is an element of order $5$, use contradiction.

Assume by contradiction there is no element of order $5$. Then, by Lagrange theorem, every element has order $1,2$ or $4$.

The problem is that $S_5$ doesn't contain soo many elements of those orders, which commute with eachother:

Any element $x \in H$ must be one of:

• a transposition
• the product of two transpositions
• a four cycle

Now, if you pick any permutation of this type, you can show it doesn't commute with 19 other permutations of this type.

Answer 2

Such a subgroup would have to contain an element of order $5$, wlog $a = (1\,2\,3\,4\,5)$, and an element $b$ of order $2$. These do not commute: Indeed, $b$ has at least one fixpoint $x\in\{1,2,3,4,5\}$ (because $5$ is odd) and for a suitable power of $a$, $a^kx$ is not a fixpoint of $b$ and also not of $a^{-k}ba^k$. We conclude $a^kb\ne ba^k$.