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Message from discussion Surjectivity of epimorphisms of groups

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More options Mar 3 2000, 3:00 am
Newsgroups: sci.math
From: magi...@yuban.berkeley.edu (Arturo Magidin)
Date: 2000/03/03
Subject: Re: Surjectivity of epimorphisms of groups
In article <38c03c2...@news.arcor-ip.de>,

Asmodeus <asmod...@arcormail.de> wrote:
>Would someone please explain to me why epimorphisms of groups are
>surjective?

Because every subgroup of a group is an equalizer subgroup.

If you want to invoke powerful theorems, you can take Schreier's
theorem on the existence of free products with one amalgamated
subgroup.

But a much easier proof is found in Carl Linderholm's "A Group
Epimorphism is Surjective", in the American Mathematical Monthly 77,
pp. 176-177:

Let f be an epimoprhism from a group G to a group H, and let A be the
image subgroup. We must show that A=H. To do this, construct two
homomorphisms g and k from H to a group K such that gf=kf (morphisms
being written on the left), and use the resulting equation g=k to show
that A=H.

Let H/A be the set of all right cosets of A. Let A' be something which
is not a right coset of A, and let S be the set H/A\cup {A'}. Let K be
the group of all permutations of S. If h, h_1, and h_2 are in H, and
if Ah_1 and Ah_2 are the same coset, then A(h_1h) and A(h_2h) are the
same coset, so the function from A/H to A/H that sends Ah' to A(h'h)
is well defined. It is easily seen to be bijective, its inverse given
by Ah'--> A(h'h_{-1}). If A' is setn to itself, this defines a
bijection of S, which we write as g(h). The function g from H to K
defined this way is actually a homomorphism (excercise left to the
reader). Lets be the permutation of S that interchanges A and A' and
leaves everything else fixed, and if h=in H, write k(h) for the
composite permutation obtained by conjugating g(h) by s. Then the
function k is the composition of g with an inner automorphism of K, so
it is also a homomorphism from H to K.

If a is an element of A, then g(a) leaves both A and A' fixed, and s
leaves every other element of S fixed, so s and g(a) commute, so
g(a)=k(a). Since g and k agree on the range of f, gf=kf. Since f is an
epimorphism, g=k.  So g(h) commutes with s for each element h of
H. Since g(h) leaves A' fixed and s exchanges A and A', it follows
that g(h) leaves A fixed. But g(h) sends A to Ah, so h lies in
A. Therefore, A=H, as desired.

Note that this also shows that an epimorphism of ->finite<- groups is
surjective (in the category of all finite groups), which does not

A Theorem of Peter Neumann shows that in any full subcategory C of the
category of all groups in which (a) every group is solvable, and (b) C
is closed under quotients; all epimorphisms are surjective. This was
extended substantially by Susan McKay. On the other hand, from an
example of B.H. Neumann and other work it is easy to exhibit full
subcategories (in fact, varieties) in which not all epimorphisms are
surjective. For example, in the variety generated by A_5, the
immersion A_4->A_5 is an epimorphism.

======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magi...@math.berkeley.edu