# Groups of order $56$

Let $G$ be a group of order $56$. (We do NOT assume the Sylow-$7$ subgroup to be normal.) Then either the Sylow-$2$ subgroup is normal or the Sylow-$7$ subgroup is normal. How to prove?

My idea: consider the case $n_2=7,n_7=8$. Then the number of elements in all Sylow $2$ -subgroups and Sylow $7$-subgroups is at most exactly $(7-1)*8+(2-1)*7+1=56$. This cannot give a contradiction. We need to prove that $G$ is simple. Then $|G||(8!/2)$. To prove $G$ simple, we need to prove there does not exist normal subgroups of order $2$ and order $4$. But I do not know how to prove.

• $\bigcup_{P_8\in Syl_2(G)}P_8\bigcup_{P_7\in Syl_7(G)}P_7\subseteq G$, and count the number of elements on both sides, we get inequality. – Shiquan Dec 11 '13 at 8:52
• what inequality? – user87543 Dec 11 '13 at 8:52
• do you feel that there is a possibility that two sylow subgroups may intersect non trivially? – user87543 Dec 11 '13 at 8:53
• ya... it is possible I think – Shiquan Dec 11 '13 at 8:56
• so,, now what do you want to do? – user87543 Dec 11 '13 at 8:56

Suppose the 7-sylow subgroup is not normal, then $n_7 = 8$, and so there are $48$ elements of order 7 in your group. This leaves exactly 8 elements which are not of order 7, and hence there is a unique subgroup of order 8 - which must be normal.

• Why must these 8 elements form a subgroup? – lhf Dec 11 '13 at 9:25
• Because if $P$ is a Sylow 2-subgroup (such group exists), its order is $8$, so $P$ is a subset of these $8$ elements and hence equal to this set! – Nicky Hekster Dec 11 '13 at 9:55

Assuming $n_7=8$, You might be convinced by now that there are $48$ non identity elements.

each sylow $2$ subgroup has $7$ non identity elements.

Suppose we have two distinct sylow $2$ subgroups $S_1,S_2$.

$S_1$ for sure will have $7$ non identity elements.

As $S_2$ is distinct form $S_1$ there would be atleast one non identity element which is not in $S_1$.

So... ? $48+7+1=??$

Do you see some thing is wrong?