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I have a proof from notes but I don't quite understand the bold part:

Abelian case: $a \in G / \{1\}$. If $\langle a\rangle \neq G$, then we are done. If $\langle a\rangle = G$, then $\langle a^4\rangle $ is a proper normal subgroup of $G$. General case: WLOG we can assume $G \neq Z(G)$. $\langle 1\rangle \neq Z(G)$ which is a proper normal subgroup of $G$. Done. Otherwise $|Z(G)|= 1$. $$ 28 = 1 + \sum_{**}\frac{|G|}{|C_G(x)|} $$ There must be some $a\in G$ such that 7 does not divide $$ \frac{|G|}{|C_G(a)|} $$ It follows that $\frac{|G|}{|C_G(a)|} = 2 $ or $4$ $\Rightarrow [G:C_G(a)] = 2$ or $4$ $\Rightarrow 28 \mid2!$ or $28\mid4!$.

Therefore group of order 28 is not simple.

Why are they true?

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up vote 4 down vote accepted

Your first displayed equation couldn't hold if all terms in the sum were divisible by $7$, since then the right-hand side would have residue $1\bmod7$ whereas the left-hand side is divisible by $7$.

Since $|G|=28$, each term in the sum must divide $28$. The only divisors of $28$ not divisible by $7$ are $1$, $2$ and $4$. The only term that's $1$ has already been taken outside the sum; if any other term were also $1$, the centre of $G$ wouldn't be trivial. Thus at least one term must be $2$ or $4$.

I don't understand what the notation at the end of the last line is intended to mean.

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You mean 28 | 2! or 4!? I think it's just the factorials. – Mark Feb 13 '12 at 6:01
@Mark: Yes -- does that mean "28 divides $2!$ or $28$ divides $4!$"? Your entire exposition is very dense and looks more like you just copied your notes rather than formulating something as a question for others to read and easily understand. – joriki Feb 13 '12 at 6:03
yes it means divides. I think this is a proof by contradiction – Mark Feb 13 '12 at 6:03
@Mark: I know that $\mid$ means "divides". The problem is that $28\mid2!$ is an assertion and $4!$ is a number, and usually numbers and assertions can't be combined using "or". What you meant would more usually be written as "$28\mid2!$ or $28\mid4!$". – joriki Feb 13 '12 at 6:09
For proofs of the result needed for that last step, see here and here. – joriki Feb 13 '12 at 6:10

There must be some $a\in G$ such that 7 does not divide $$ |G| \over |C_G(a)| $$

For the first thing, note that LHS is divisible by $7$. If all the terms in the summation were divisble by $7$ then, the RHS, will be $1 \mod 7$. So, there must be atleast one $x$ with the property you claim.

It follows that ${|G| \over |C_G(a)|} = 2 $ or $4$

Note that, $I=\dfrac{|G|}{|C_G(x)|}$, called the index of $C_G(x)$ in $G$, is a number that divides the order of the group. (Why?)

So, if $I$ is not $7$, it must be $2$ or $4$. (It cannot be $1$ by the very definition of class equation.)

$\Rightarrow [G:C_G(a)] = 2$ or $4$ $\Rightarrow 28 |2!$ or $4!$.

The conclusion in the last line is a very important lemma due to Poincare. Let $G$, a simple group act on $G/H$, where $H$ is a proper subgroup of $G$. This gives you a homomorphism from $G \to \operatorname{Sym}(G/H)$. Is the action trivial? Now what will its Kernel be? What does the first isomorphism theorem tell you?

P.S.: I quite disagree with the sequel, especially in the last part. You start with the assumption that group is simple. Then you arrive at a contradiction. *"Therefore group of order 28 is simple." * is placed wrongly in your proof.

A nicer way would be to:

Suppose $G$ is a simple group. Then, its centre is trivial.

Because, if this was non-trivial, this will be a non-trivial normal subgroup of $G$. If it is proper, we are done because this is a contradiction to the "Simpleness" of the group. If it is improper, then, the group is abelian. By Cauchy's Theorem for Abelian groups, there must be an element of order $2$ which should be normal (because any subgroup in $G$ will be normal).

Now carry out the argument I have showed you almost explicitly!

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@Mark - or you could skip the argument based on the class formula and apply Sylow's theorems: the 7-Sylow subgroup must be normal, since the number of such subgroups must divide 28 and must be congruent to 1 (mod 7). Probably at this point in your notes the Sylow theorems haven't crossed your path ...

Probably more basic: by Cauchy's theorem, $G$ has a subgroup of order 7, say $H$. Now $G$ acts on the four left cosets of $H$, by left multiplication. The kernel of this action is the normal subgroup $core_G(H)$, and is contained in $H$, hence it equals $H$ or must be trivial (its order divides the prime 7). But $|G/core_G(H)|$ must divide $4!$ (this is hopefully one of the theorems you learned already). Hence $H = core_G(H)$, so $H$ is normal, whence $G$ not simple.

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The @ thing in an answer doesn't ping people. On the other hand, answering someone's question pings them automatically. – anon Feb 28 '12 at 23:39
"since the number of such subgroups must divide 28". I think it should be "must divide 4" – yoyostein Jul 8 at 7:26
@yoyostein :-)) I pointed out that the number is congruent $1$ mod $7$, hence relatively prime to $7$, so this implies your remark. The number of Sylow $p$-subgroups always divides the order of the group. – Nicky Hekster Jul 8 at 7:47
I see. Thanks!!! – yoyostein Jul 8 at 7:55

Reading the class equation modulo 7 gives the existence of one $x$ such that $\frac{|G|}{|C_G(x)|}$ is NOT divisible by 7. Hence 7 divides $|C_G(x)|$. Now the factors of the numerator $|G|$ are 1, 2, 4, 7 $\cdots$. Since $\frac{|G|}{|C_G(x)|}$ cannot be 1 and cannot divide 7, the only possibilities are 2 and 4.

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