NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/sci/math/17468

Internet Message Format | 1992-12-28 | 5.2 KB

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From: [e-mail] (Frederick W. Chapman)
Newsgroups: sci.math
Subject: H < G finite abelian ==> G/H ~= K for some K < G
Message-ID: <[e-mail]>
Date: 28 Dec 92 16:06:07 GMT
Organization: Lehigh University
Lines: 96

In article <[e-mail]>,
[e-mail] (David Bradley) writes:

>This has probably been discussed before, but could someone please
>nudge me in the right direction on this one?  It's not homework.
>
>Let G be an abelian group and H a subgroup.  Then G has a subgroup
>isomorphic to G/H.  I managed to prove this using the isomorphism
>G ~ Hom(G,Q/Z), but surely there is a more direct proof, perhaps
>based on the structure theorem for finite abelian groups.  This is
>embarassing.

As you mentioned in your second post (not quoted here), you are working
over *finite* abelian groups.  (The proposition is false for some infinite
groups; e.g., Z has no subgroup isomorphic to Z/2Z since every non-zero
element of Z has infinite order).

Here is an outline for a proof based on the fundamental theorem for finite
abelian groups.  Write all groups additively and let (+) denote direct sum.
Let ~= denote isomorphism.  If G is any group, write H < G if H is a
subgroup of G, and let |G| denote the order of G.

Using the fundamental theorem, we can write G = G_1 (+) ... (+) G_n, where
each G_i < G is cyclic of prime power order, for i = 1, ..., n.  Note that
we elect to use the internal rather than external direct sum, which allows
us to write this as an equality rather than an isomorphism.  The part of
this proof which takes some thought is to show (by applying the fundamental
theorem to H) that we can write H = H_1 (+) ... (+) H_n, where each H_i < G
is also cyclic of prime power order, and, in addition, |H_i| divides |G_i|,
for i = 1, ..., n; note that some or all of the H_i's may be the trivial
subgroup {0}.  Since both G_i and H_i are cyclic, and |H_i| divides |G_i|,
there exists J_i < G_i such that H_i ~= J_i, for i = 1, ..., n.  Let J =
J_1 + ... + J_n.  Since J_i < G_i for each i, and the sum of the G_i's is
direct, the sum of the J_i's is also direct, and we can in fact write J =
J_1 (+) ... (+) J_n.  Since H_i ~= J_i for i = 1, ..., n, it follows that H
~= J.  Thus, although it is NOT NECESSARILY possible to write an arbitrary
subgroup H as the direct sum of subgroups of the G_i's (see counterexample
below), it is always true that H is ISOMORPHIC to a subgroup J which CAN be
written as the direct sum of subgroups of the G_i's.  It now follows that

                G/J ~= (G_1/J_1) (+) ... (+) (G_n/J_n).

Since each G_i is cyclic, there exists K_i < G_i such that G_i/J_i ~= K_i,
for i = 1, ..., n.  Since the sum of the G_i's is direct, the sum of the
K_i's is direct also, and we can define K < G by K = K_1 (+) ...  (+) K_n.
We conclude that G/H ~= G/J ~= K.

...........................................................................

If the finite abelian group G is cyclic, then the subgroup K such that G/H
~= K is actually unique.  The existence and uniqueness of K for cyclic
groups G follow from these facts: (1) every finite cyclic group G has
exactly one subgroup of order d for every positive integer d which divides
the order of G; (2) every quotient of a cyclic group is cyclic; (3) every
subgroup of a cyclic group is cyclic; (4) two cyclic groups of the same
order are isomorphic.

If G is not cyclic, the subgroup K need not be unique, as the following
example shows.  Define a finite additive abelian group G by generators and
relations, thus:

                  G = < a,b | 4a = 4b = 0, a+b = b+a >.

Let C_n denote the cyclic group of order n.  Clearly, G ~= C_4 (+) C_4.
Define H < G by H = < 2a, b >; then H ~= C_2 (+) C_4 and G/H ~= C_2.  We
can define subgroups K = < 2a > and K' = < 2b >.  It follows that K ~= K'
~= C_2 ~= G/H; however, K and K' are distinct subgroups of G.

We can use this group G to construct the counterexample referred to in the
proof above.  Define subgroups G_1 = < a >, G_2 = < b >; then G = G_1 (+)
G_2, where the G_i's are cyclic of prime power order.  Now define subgroup
H = < a+b >.  H cannot be written as H = H_1 (+) H_2, where H_i < G_i for
each i; however, H ~= J = J_1 (+) J_2, where J_1 = G_1 and J_2 = {0}, which
satisfies J_i < G_i for each i.

...........................................................................

I hope that this outline is useful.  The result is a nice application of
the fundamental theorem of finite abelian groups and facts about finite
cyclic groups.

Sincerely,


Frederick W. Chapman
Senior User Consultant
Lehigh University Computing Center

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