Re: ALL the groups of order 24

From: Van www (vanjac12_at_yahoo.com)
Date: 11/25/04 

Date: 25 Nov 2004 04:02:02 -0800

Dic(C_6) x C_2 = C_6 x| C_4 ; I don't know which of Pawel Gladki's
groups this corresponds to, but x^6 = 1 = y^4, with yxy^(-1) = x^(-1)
is the only possible non-Abelian group. Conjugation by y fixes x^3,
and y^2 x = x y^2, so Z = <x^3> x <y^2> = C_2 x C_2, and G/Z = D_3.

The last thing is to verify that

D_4 x C_3 = C_3 x (C_4 x| C_2) = C_4 x| C_6 = (C_2 x C_2) x| C_6

which I don't quite understand, since D_4 = C_4 x| C_2, is clear,
acting by conugation, with yxy^(-1) = x^(-1), so the action squared
is the identity.

But A(C_2 x C_2) = D_3 has actions of order both 2 and 3.

I suppose we could choose the action of order 2, interchanging
the generators of C_2 x C_2, and ignore the action of order 3,
which is the cyclic permutation of the 3 non-identity elements of
C_2 x C_2, but it does not seem like this would give a group
isomorphic to the first, though I haven't checked it.
I guess these groups must be isomorphic, as this is the last group.
If anyone has comments on this last group I would welcome them.

Van

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