How can I prove that there is no simple group of order $448=2^6\cdot 7$? I tried with Sylow's theorems, I proved that (if $G$ is simple) the number of 2-Sylows is 7 and that the number of 7-Sylows is 8 or 64, but I don't know how to continue, could you help me please?
There is no simple group of order $448=2^6\cdot 7$
3
$\begingroup$
group-theory
finite-groups
simple-groups
-
1(There is a monthly maximum for questions... :D ) – 2012-01-03
-
1Hint: If there are 64 Sylow 7-subgroups, consider how may elements of order 7 there are. – 2012-01-03
-
0@GeoffRobinson: $64\cdot 6=384$ elements, right? How can I continue? – 2012-01-03
-
0@MarianoSuárez-Alvarez: really? How many questions can I ask? – 2012-01-03
-
2@Alex: Well, if there are 384 elements of order $7,$ how many Sylow $2$-subgroups can there be? (Actually, I see a more direct approach to the question than this anyway). – 2012-01-03
-
2only 1, right thank you – 2012-01-03
-
0Is there a way to show the Sylow 7-subgroup is normal that doesn't prove the group is not simple long before? In general, the Sylow $p$-subgroup in a group of order $q^n p$ cannot normalize a non-identity $q$-subgroup unless it normalizes the entire Sylow $q$-subgroup, but even in the $q=2$, $p=7$ case one immediately gets a non-identity normal 2-subgroup from the assumption of 7 Sylow 2-subgroups. – 2012-01-03
-
0@Alex: No, it might be $7.$ – 2012-01-03
-
0@ehsanmo: if there are more than 1 sylow 7-subgroups, then there must be exactly 1 sylow 2-subgroup. – 2012-01-03
-
0@JackSchmidt: right! I'd partially read the comment. – 2012-01-03
-
0@JackSchmidt: I'm not sure I understand your claim about groups of order $q^np$. If $P$ is any 2-group, you seem to be saying that the 7-subgroup of the dihedral group cannot normalizes it in $P\times D_7$? – 2012-01-03
-
0If we look at a putative (ok, non-existent by Burnside $q^ap^b$) simple group of order $2^n\cdot7$, won't the argument starting from the study of a maximal intersection of two Sylow 2-subgroups always lead to the conclusion that all Sylow 2-subgroups need to intersect trivially. This might actually work for all $q,p$? – 2012-01-03
-
1@Jack, Jyrki: I do not believe the statement about groups of order $q^{n}p.$ For example, the symmetric group $S_4$ has order $2{3}.3,$ and a Sylow $3$-subgroup normalizes the normal Klein $4$-group, but not a whole Sylow $2$-subgroup (there is noting special about these primes for this question). – 2012-01-03
-
0There is a typo above : it should of course be $2^{3}.3$. – 2012-01-03
-
0@Geoff, Jyrki: Thanks for the correction. In the light of the morning, my proof only shows $G/O_q(G)$ has a normal Sylow $p$-subgroup, which is pretty obvious from *G* being $p$-solvable. There is a group of order $2^7⋅7$ with no normal Sylow subgroups, and a similar group exists for $q=2$ and every *p* (take a faithful irreducible GF(2) module for *p*, then let *p* act on it and its dual, then another copy of 2 switches them), and its not hard for me to believe something similar works for all *q*. Even simpler: take a faithful $D_{2p}$ module. – 2012-01-03
-
0@Jack: Yes, let a group $Q$ of order $q$ act faithfully and irreducibly on an elementary Abelian $p$-group $P.$ Then let $QP$ act irreducibly and faithfully on an elementary Abelian $q$-group $V$. The semidirect product V(PQ) has neither a normal Sylow $p$- nor a normal Sylow $q$-subgroup. This is just what goes on with $S_4,$ which is the case $q=2,p=3$. – 2012-01-03
-
0$7\not\equiv 1\pmod{4}$, so if there is more than one Sylow 2-group, then there exist two Sylow 2-groups $P$ and $Q$ such that $P\cap Q$ has order 32, and subsequently must be normal. Similar reasoning shows that for $m>1$, a group of order $2^m\cdot7$ always has nontrivial $O_2$. – 2012-04-16
1 Answers
6
Let $n_2$ be the number of $2$-Sylow subgroups of $G.$ Then, $n_2$ is odd and divides $7.$ If $n_2=1$ we're done, but if $n_2=7$, by Sylow theorem, conjugation of these seven $2$-Sylow subgroups defines a homomorphism $G \to S_7.$ The kernel of this homomorphism cannot be trivial so we're done.
-
0why the kernel cannot be trivial? – 2012-01-03
-
3+1: This works for groups of size $2^6\cdot 7$, because that number does not divide $7!$. But curiously it wouldn't work for $2^4\cdot 7$ :-) – 2012-01-03