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Prove that there are no simple groups of order 224.

Let $G$ be a finite group such that $\vert G \vert = 224 = 2^5 \cdot 7$. We know that $n_2 \mid 7$ and $n_2 \equiv 1 \pmod 2$ and we know that $n_7 \mid 2^5$ and $n_7 \equiv 1 \pmod 7$. So we can say $n_2 = 1$ or $7$ and $n_7 = 1$ or $8$. Suppose, to the contrary that $G$ is a simple group. Then $n_7 = 8$ and $n_2 = 7$. So we can say there are $8 \cdot 6 = 48$ elements of order 7 and $7 \cdot 31 = 217$ elements of order 2, which gives us 265 elements, 266 including the identity, which contradicts the cardinality of the group. Hence, $G$ is not a simple group since we must have at least one of $n_7$ or $n_2$ being 1.

Is this approach correct? specifically how i said there were $7 \cdot 31$ elements of order 2? I saw a few similar approaches online and thought it would work nicely for this problem. Is it correct?

Thank you!

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Yep, there seems to be a problem with saying there are $7 \cdot 31$ elements of order $2$. If I understand your notation correctly, $n_2=7$ is the number of Sylow $2$-subgroups. Each such subgroup has order $32$. But elements of such a subgroup can have orders other than $1$ and $2$: in general, there can be larger powers of $2$ there too. And even if there are $31$ elements of order $2$ in each Sylow $2$-subgroup, you also implicitly use that any two such subgroups have a trivial intersection, and It's not clear why that should be true. – Dan Shved Nov 7 '12 at 11:48
I thought for each prime $p$ dividing $\vert G \vert$, the Sylow $p$-subgroups intersect trivially? – Robert Cardona Nov 7 '12 at 11:52
Nope, that's not necessary at all. For example, you can look at this question (I suspect that it originated from the same problem):… – Dan Shved Nov 7 '12 at 12:01
Ok, my bad, I'll keep working on it! – Robert Cardona Nov 7 '12 at 12:03
up vote 5 down vote accepted

The following proves there must exist a normal Sylow $\,2-$subgroup of order $\,32\,$,

Suppose there are $\,n_2=7\,$ Sylow $\,2-$subgroups in $\,G\,$ . Making $\,G\,$ act on the set of these Sylow subgroups by conjugation (Mitt wrote about this but on the set of the other Sylow subgroups, which gives no contradiction), we get a homomorphism $\,G\to S_7\,$ which must be injective if $\,G\,$ is simple (why?).

But this cannot be since then we would embed $\,G\,$ into $\,S_7\,$ , which is impossible since $\,|G|\nmid 7!=|S_7|\,$ (why?)

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Since it is not injective, it has a nontrivial kernel, that's normal in $G$, so its not simple? – Robert Cardona Nov 7 '12 at 12:32
Yup, exactly. I'd rather say: "Since it can't be injective then that homomorphism's kernel is non trivial and thus $\,G\,$ is not simple"...and in fact we proved a little more: that one of the group's Sylow subgroups is normal. – DonAntonio Nov 7 '12 at 12:35
I totally agree with the approach of DonAntonio ... and Robert, probably you have heard of this famous theorem of Burnside that every group of order $p^aq^b$ is solvable? But the digging into the Sylow subgroups and doing the analysis for particular cases is always fun to do. – Nicky Hekster Nov 7 '12 at 13:07
@Nicky Thanks I'll do that. My professor 'suggested' we look at finite groups of order 1-60 and play around with them to learn as much as we can. Thanks! – Robert Cardona Nov 7 '12 at 20:32
To complete your (why?) questions: The mapping must be injective because if it is not, the kernel would be non-trivial, and since the kernel is a normal subgroup of $G$ it would imply $G$ is not simple; $G$ must divide the order of $S_7$ because if it was injective, it would map bijectively to a subgroup of $S_7$ meaning it was isomorphic to a subgroup of $S_7$ meaning it would have to divide the group? – Robert Cardona Nov 8 '12 at 0:36

That part is false. From $n_7=8$, you have eight subgroups of order 7. Each of these subgroups is cyclic of prime order, so the six nontrivial elements must be of order 7, hence giving $8\cdot 6=48$ elements of order 7. (You also need to check they are disjoint.) But for the Sylow 2-subgroup, they are of order 32. There is no reason why the nontrivial elements are of order 2.

Hint to another approach: Consider the conjugation action of $G$ on the eight Sylow-7 subgroups. This gives a injective homomorphism from $G$ to $S_8$.

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There are even less $2$-Sylows ($G \to S_7$). [I guess you mean injective in the last sentence.] – j.p. Nov 7 '12 at 12:13

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