There is no simple group of order 56

Question

I'm working on question 16 of http://www.austinmohr.com/Work_files/hw4.pdf

I'm confused by why there are 48 elements of order 7. I understand that $n_7 = 8$, but no further.

Answer 1

In general, if $G$ is a group of order $p^3q$, where $p$ and $q$ are primes such that $p \lt q$, then $G$ is not simple.

Proof: if $Q \in Syl_q(G)$, then $|G:N_G(Q)|=1, p, p^2$ or $p^3$. If $n_q=1$, then $Q$ is normal and we are done. If $n_q=p$, then $|G:N_G(Q)|$ is the smallest prime dividing the order of $G$, and it is well-known that this implies that $N_G(Q) \lhd G$. If $n_q=p^3$, then we can apply a similar counting argument as above: the $p^3$ Sylow $q$-subgroups contain a total of $p^3(q-1)$ elements of order $q$ and hence $G$ contains at most $p^3q-p^3(q-1)=p^3$ elements that do not have order $q$. It follows that a Sylow $p$-subgroup must be normal in this case.

So we are left with the case where $n_q=p^2$. But then $p^2 \equiv 1$ mod $q$. This implies that $q \mid (p-1)(p+1)$ and since $q$ is prime, we get $q \mid p-1$ or $q \mid p+1$. So in any case $q \leq p+1$. Since $p \lt q$, the only possibility is that $p+1=q$. But then $p=2$ and $q=3$, and $|G|=24$. It is well-known (and left as an exercise) that groups of order $24$ are not simple.

Answer 2

We have that $56=2^3\cdot 7$. Then the number of Sylow $7$-subgroups is $1+7k$ , for some $k \geq 0$ and divides $8$. So the number of Sylow $7$-subgroups is either $1$ or $8$. If it is $1$, we are done. So let us assume that it is $8$. Each Sylow $7$-subgroup has order $7$. In a group of prime order every nonidentity element has order $7$. Thus each Sylow $7$-subgroups has $6$ elements of order $7$. Also, they intersec trivially. It follows that there are $8\cdot 6 =48$ elements of order $7$. This leaves only $56-48=8$ elements, wich is the size of one Sylow $2$-subgroup. Thus, there is only room for one o such subgroups. That is, either the Sylow $7$-subgroup or the Sylow $2$-subgroup must be normal.

Answer 3

In general, if $G$ is a group of order $pq$, where $p$ and $q$ are (not necessarily different) primes, then $G$ is not simple.

Proof: if $p=q$, then the order of $G$ is a square of a prime and it is well-known that such a group is abelian. But simple abelian groups have prime order. So assume that $p \neq q$. Let $Q \in Syl_q(G)$. If $Q$ is normal we are done. Hence let $Q$ be not normal, that is $N_G(Q) \lt G$. Then, since $Q \subseteq N_G(Q) \subset G$ and $|G:Q|=p$, we see that $Q=N_G(Q)$ and we conclude that $n_q=p$.

Observe that each pair of Sylow $q$-subroups intersect trivially: if $Q, R \in Syl_q(G)$, with $Q \neq R$, then $Q \cap R$ is a subgroup of the group $Q$, which has prime order $q$. By Lagrange, $Q \cap R=1$ or $Q \cap R=Q$. In the latter case we have $Q=Q \cap R \subseteq R$, whence $Q=R$ contradicting the assumption.

Now let us count the number of elements of order $q$: each $Q \in Syl_q(G)$ contains $q-1$ elements of order $q$. Hence, there are in total $n_q(q-1)=p(q-1)$ different elements of order $q$. That leaves $pq-p(q-1)=p$ elements of order $p$. But by Sylow there must be at least one $P \in Syl_p(G)$. So all the elements of order $p$ must form this $P$. Since there are no more $p$ elements to be found it means that there is only one Sylow $p$-subgroup, implying (all Sylow subgroups are conjugate), $P$ is normal. And we are done.