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So $280 = 2^3\cdot 5 \cdot 7$. I assume $G$ is simple. I'm having trouble using the Sylow Theorems to show that this is not Simple. In particular, computing the number of sylow groups and using that to show the group isn't Simple.

Algnoe
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  • What possible numbers did you come up with for the various Sylow subgroups? – Tobias Kildetoft May 06 '15 at 09:49
  • Well, for $n_2$, we would have $n_2 \equiv 1 mod 3$ and $n_2 \mid 45$ right? Then $n_2 = 7$ is the only number that satisfies this. But then for $n_5$, I think I am messing up. I say $n_5 \equiv 1 mod 1$ and $n_5 \mid 56$. But $1 mod 1$ does't make any sense – Algnoe May 06 '15 at 10:00
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    You are doing mod the wrong thing. It should be mod the given prime, not mod the exponent. – Tobias Kildetoft May 06 '15 at 10:01
  • Ah, whoops. So $n_2 \equiv 1 \mod{2}$ and $n_2 \mid 45$ which means $n_2 = {1,5,7}$. $n_5 \equiv 1 \mod{5} $ and $n_5 \mid 56$, but there are no $n_5$ which satisfies this except $1$. For $n_7 \equiv 1 \mod{7}$ and $n_7 \mid 40$ then $n_7 = {1,8}$. – Algnoe May 06 '15 at 10:19
  • @Algnoe Where in the world are you taking that poor $;45;$ from... again ?? – Timbuc May 06 '15 at 10:20
  • Multiply is hard, $n_2 \equiv 1 \mod 2$ and $n_2 \mid 35$ which mean $n_2 = {1,5,7}$ – Algnoe May 06 '15 at 10:22
  • @Algnoe In fact, $;n_2\in{1,,5,,7,,7\cdot5=35};$ . I hope your remark on multiplication was jesty... – Timbuc May 06 '15 at 10:26

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Suppose the group isn't simple $\;\implies n_5=56\implies\;$ there are $\;56\cdot 4=224\;$ elements of order $\;5\;$

Also, we have that $\;n_7\ge8\implies\;$ there are at least $\;8\cdot 6=48\;$ elements of order $\;7\;$ .

Thus, we already have $\;224+48=272\;$ elements of order $\;5\,\,\,or\;\;7\;$ . The ones left must belong to the then unique Sylow $\;2$-subgroup ...

Other proof's highlights . Suppose $\;n_5=56\;$ , then there exists a subgroup of $\;G\;$ with this order and this with index $\;\frac{280}{56}=5\;$ (namely, any Sylow $\;5$-subgroup's normalizer).

But then the action of $\;G\;$ on this subgroup's left cosets (i.e., the regular left action) renders a homomorphism $\;G\to S_5\;$ which, if $\;G\;$ is simple, must be injective. But this is absurd as $\;|G|\nmid 5!=120\;$ .

Timbuc
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