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Show that any group of order 8 is not a simple group. I know that $\mathbb{Z}_8$, $\mathbb{Z}_2\times \mathbb{Z}_4$, $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$, $Q_8$, $D_4$ are not simple. But I am unable to prove it generally.

Please help.

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Hint: If all elements have order $2$ the group is abelian. If there is an element of order $4$, it generates a subgroup of index $2$, which is then normal. –  Tobias Kildetoft Apr 30 '13 at 15:02
Here is a silly answer that is wrong: The number of Sylow 2-subgroups is a divisor of 8 that is congruent to 1 mod 2, so the number of Sylow 2-subgroups is exactly 1. Since there is only one Sylow 2-subgroup, it must be normal. $\square$ –  Jack Schmidt Apr 30 '13 at 15:06
@JackSchmidt Very nice. niladri: Note that the method I used here was intentionally very elementary. In general, one can show that no group of order $p^n$ for a prime $p$ is simple (they are in fact nilpotent), but this requires a bit more argument. –  Tobias Kildetoft Apr 30 '13 at 15:10
You can also use the classification of simple groups and check that no group in that list has order $8$.... Just kidding, sorry couldn't resist.... –  N. S. Apr 30 '13 at 15:20
I should perhaps add that what I wrote about groups of order $p^n$ is not completely correct, as any group of prime order is of course simple by Langrange's theorem. –  Tobias Kildetoft Apr 30 '13 at 15:22

2 Answers 2

Here is another silly proof, but this one is probably right.

By Cauchy's theorem (or Lagrange) a group $G$ of order 8 contains a subgroup $H$ of order 2. Consider the homomorphism from $G$ to $\operatorname{Sym}(4)$ given by number the 4 cosets of $H$ in $G$, and letting $G$ act as multiplication on the cosets. The image is transitive (moves all the points around) so the kernel must be smaller than $G$. If $G$ is simple, then the kernel has to be the identity. The first isomorphism theorem shows that $G$ is isomorphic to a subgroup $\operatorname{Sym}(4)$, but every subgroup of order 8 in $\operatorname{Sym}(4)$ is a Sylow 2-subgroup, and so dihedral of order 8. Since dihedral groups of order 8 are already known to be non-simple, we are done. $\square$

Usually instead of the cosets of a subgroup of order 2, one uses a bigger subgroup, but 8 is so small, this just works anyways.

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Note that Tobias's hints lead to a simpler proof. This proof is more suited to "prove there is no simple group of order 24" (H is taken to have order 3, and one concludes the simple group is S4, which isn't simple, oops). –  Jack Schmidt Apr 30 '13 at 15:27

This group is a $p$-group with $p=2$, so the center of the group, $Z(G)$, is nontrivial, and $Z(G)$ is normal in $G$ (true for any group). Moreover, $Z(G)=G$ $\iff$ $G$ is abelian, in which case any subgroup of order $2$, for this specific problem, is normal. If $Z(G)\neq G$, then $Z(G)$ is a nontrivial proper normal subgroup.

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