Re: trouble classifying gro.ups of order 56

magidin_at_math.berkeley.edu
Date: 12/15/04 

Date: 15 Dec 2004 14:44:15 -0800

Jorge Caliva wrote:
> >What are you doing?
>
> sorry, I had a large notational mix up (I started refrencing the
> book, in which there is an example where a,b are generators of
> the group being acted on).
>
> I understand your work, I just didn't see that the actions could
> be seen to be equivalent because of an automorphism exchanging a
> and b.

Say that a group G acts on a group K, and that f:G->G is a group
automorphism.

The semidirect product corresponds to an action; and an action
coresponds to a group homomorphism a:G-> Aut(K).

Now, since f is a group homomorphism (in fact, an automorphism), we get
a new homomorphism b:G-> Aut(K) via

b(g) = a(f(g)) (i.e., compose with f).

The claim is that the two actions yield isomorphic extensions.

Remember that the semidirect product can be realized as the group with
underlying set K x G, and with multiplication given by

(k,g) * (h,m) = (k h^g, gm)

where h^g is the image of h under the action of g. If we use, it is h^g
= a(g)(h); if we use b, it is b(g)(h) = a(f(g))(h).

I claim the two semidirect product we obtain are isomorphic. The
isomorphism will be given by sending the pair (k,g) (when the action is
given by b) to the pair (k,f(g)) (when the action is given by a).

First, this is clearly a bijection of underlying sets, since f is an
automorphism of G.

And it is a group homomorphism: for in the action given by b,

(k,g) * (h,m) = (k*b(h), gm)

The image of the right hand side under our map is

(k*b(h), f(gm) )

The images on the left hand side are

(k,f(g)) and (h,f(m)). When we multiply them together using the action
a, we get

(k,f(g)) * (h,f(m)) = (k*a(f(g))(h) , f(g)f(m) )

Since f(g)f(m) = f(gm) and since b(g) = a(f(g)), we see that the image
of the product is the product of the images, so f is indeed an
isomorphism.

Now consider your example: Q_8 is generated by a and b; you have two
actions: one in which a acts trivially and b acts by inversion; and one
in which a acts by inversion and b acts trivially. Note that the second
action is equal to composing an automorphism of Q_8 that exchanges a
and b with the first action. Thus, the two actions yield isomorphic
extensions.

For your third action, where a and b both act by inversion, note that
Q_8 is generated by a and ab; in your first action, a acts trivially
and ab acts by inversion; here, a acts by inversion and ab acts
trivially. Since there is an automorphism of Q_8 that swaps a and ab,
that means the two actions again yield equivalent group extensions.

(This is not the only way in which two extensions can be isomorphic)

> r.x = x s.x = x
> r.x = x^{-1} s.x = x
> r.x = x s.x = x^{-1}
> r.x = x^{-1} s.x = x^{-1}
>
> Now, since these four actions give rise to only 3 non-isomorphic
> groups, 2 of hem must be equivalent. unfortunately, i still do
> not understand how I can determine this. My only guess was
> that possibly the 4th action was the trivial action, but writing
> out the group law I do not think this is true:

It isn't. It cannot possibly be. Because in the trivial action, every
element of D_8 acts trivially on every element of Z_7; but in the other
three actions, there is at least one elmeent of D_8 that does not act
trivially on at least one element of Z_7; the action cannot be trivial
and the extension cannot be equivalent to the trivial action. Two of
the latter three nontrivial actions yield equivalent extensions.

I assume that your r and s in the generation of D_8 are the ones that
satisfy

r^4 = s^2 = 1, sr = r^3s

Is this correct?

(I ask, because D_8 can be generated by two elements of order 2, and it
might be easier to see then what's going on; say, generate it with s
and rs).

Arturo Magidin, sans.sig

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