# $7$ Sylow subgroup of $A_{20}$ is not normal and $7$ Sylow subgroup of $S_{20}$ contained in $A_{20}$.

Let $$H$$ be a 7 Sylow subgroup of $$A_{20}.$$ Then prove that $$H$$ is not normal in $$A_{20}$$ and any 7 Sylow subgroup of $$S_{20}$$ is contained in $$A_{20}.$$

Proof:

since $$7^2$$ divides order of $$A_{20}$$ but $$7^3$$ does not divide, we have $$|H|=49.$$ Now there are more than $$2 \times6!=2 \times720$$ elements of order $$7$$ in $$A_{20}$$, because $$(1\ 2\ 3\ 4\ 5\ 6\ 7)$$ and $$(8\ 9\ 10\ 11\ 12\ 13\ 14)$$ are of order $$7$$ in $$S_{20}$$ and they are even permutations so contained in $$A_{20}$$. If $$H$$ is a normal 7 Sylow subgroup then it must be unique in $$A_{20}$$. And every element of order $$7$$ must be contained in $$H$$ (By Sylow's second theorem), which is not possible because $$H$$ can at most contain $$49$$ elements. So $$H$$ is not unique. Am I correct in this part?

In second part, every element of order of 7 Sylow subgroup is of order $$7$$ or $$1$$. So it must be even permutation. So it is contained in $$A_{20}$$. Am I correct? Thank you.

• I guess the result that $A_n$ is simple for $n\geq5$ is not to be used? Dec 1 '18 at 13:03
• @Servaes we can use that result. But I forgot that result while proving this. It becomes so simple using this result. Thanks Dec 1 '18 at 13:31

Your approach seems fine to me, though your first argument is a bit cluttered. They key observation you have made is that if $$H$$ is normal in $$A_{20}$$, then the elements of $$A_{20}$$ of order $$7^k$$ ($$k\in\{0,1,2\}$$) form a subgroup of order $$49$$. So indeed it suffices to show that there are more than $$49$$ elements of order $$7^k$$ in $$A_{20}$$. But I do not understand the argument you give for the existence of $$2\times6!=2\times720$$ elements of order $$7$$ in $$A_{20}$$.
My first thought would be to note that any choice of $$7$$ elements from $$S_{20}$$ yields $$6!$$ distinct $$7$$-cycles in $$S_{20}$$, yielding at least $$\binom{20}{7}\times6!>49$$ elements of order $$7$$. (There are even more though.)
Your idea for the second part is good, though the implication $$\text{ the order of \sigma is 7 or 1}\qquad\Rightarrow\qquad \sigma\in A_{20},$$ requires a (simple) argument.
• We can permute $(1234567)$ in $6!$ ways. Similarly for $(8 9 10 11 12 13 14)$. So I wrote more than $2 \times 6!$ Dec 1 '18 at 13:35