Re: Embedding an elementary abelian group into a symmetric group

On 14.09.2008 13:53, Narcoleptic Insomniac wrote:
On Sep 14, 2008 3:47 AM CT, Jannick Asmus wrote:

On 14.09.2008 09:55, Narcoleptic Insomniac wrote:
On Sep 14, 2008 12:59 AM CT, steenrod wrote:

Thanks, Kyle.

Incidentally, why is your map injective?
For two reasons:

1) Let x and y be distinct elements in H = (Z/2)^5, and let g be
an element of G = <(01),(23),...,(89)> such that phi(x) = phi(y)
= g.

Since all non-trivial elements of G have order two, it follows
that

e_G = g^2 = phi(x) phi(y)

...but phi is a homomorphism, so

e_G = g^2 = phi(x) phi(y) = phi(xy) = phi(e_G)

...which implies that y = x^{-1}.
Why that? I cannot see your argument. ;)

^_^ Are you referring to the typo?

In that case, the last string of equalities should read:

e_G = g^2 = phi(x) phi(y) = phi(xy) = phi(e_H).

If not, then allow me to break it down:

1) We begin with the identity element e_G of G.

2) Since all non-trivial elements of G are order 2, we must have that
e_G = g^2.

3) By hypothesis we're assuming that phi(x) = phi(y) = g, so g^2 =
phi(x) phi(y).

4) Since phi is a homomorphism, phi(x) phi(y) = phi(xy).

5) Again, since phi is a homomorphism, phi(e_H) = e_G; hence, e_G =
g^2 = phi(x) phi(y) = phi(xy) = phi(e_H).

In other words, because phi sends xy to the identity in G, it follows
that xy must be the identity in H...

..for, if xy were not equal to e_H, phi would be mapping two elements
to e_G and would thus not be a homomorphism.

I buy everything - but the last thing _as it is_.

More precisely: I understand that you are saying that if phi is not injective, then it cannot be a homomorphism. Why that?

I would expect here some more analysis on the groups involved, since the statement is not true in general.

This establishes that xy = e_H and implies y = x^{-1}.

Alternatively you can show that G is a vector space over Z/2Z of
dimension 5 and f is a surjective linear map (over Z/2Z), so it is
injective.

I suppose, but there is no need to induce more structure on the group
(i.e. consider it also as a vector space) in order to prove such a
group theoretical fact...

..but whatever flips your switch ^_^

Regards, Kyle Czarnecki

--
Best wishes,
J.
.