2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $\bigcup_{P_8\in Syl_2(G)}P_8\bigcup_{P_7\in Syl_7(G)}P_7\subseteq G$, and count the number of elements on both sides, we get inequality. $\endgroup$ – Shiquan Dec 11 '13 at 8:52
  • $\begingroup$ what inequality? $\endgroup$ – user87543 Dec 11 '13 at 8:52
  • $\begingroup$ do you feel that there is a possibility that two sylow subgroups may intersect non trivially? $\endgroup$ – user87543 Dec 11 '13 at 8:53
  • $\begingroup$ ya... it is possible I think $\endgroup$ – Shiquan Dec 11 '13 at 8:56
  • $\begingroup$ so,, now what do you want to do? $\endgroup$ – user87543 Dec 11 '13 at 8:56
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Why must these 8 elements form a subgroup? $\endgroup$ – lhf Dec 11 '13 at 9:25
  • 2
    $\begingroup$ 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! $\endgroup$ – Nicky Hekster Dec 11 '13 at 9:55
0
$\begingroup$

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?

$\endgroup$

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.