Re: Algebra with |HK|.

In article <f8ngk8$939$1@xxxxxxxxxxxxxxxx>,
mina_world <mina_world@xxxxxxxxxxx> wrote:
Hello sir~

If H and K are finite subgroups of a group G,
then |HK| = (|H||K|) / |H /\ K|.

-----------------------------------------------
I prepared two proof.
(1)
Let |H /\ K| = t. so H /\ K = {c_1, c_2, .... , c_t}.
Let H = {a_1, a_2, .... , a_i, c_1, c_2,.... , c_t}.
Let K = {b_1, b_2, ...., b_j, c_1, c_2, ...., c_t}.

Let A = {a_1, a_2, .... , a_i} = H - (H/\K).
Let B = {b_2, b_2, ...., b_j} = K - (H/\K).

Consider HK = A.B U A.(H/\K) U (H/\K).B U (H/\K).

Okay...

For any a in A and any c in H /\ K,
a.c not in H /\ K.
[Because, If a.c in H /\ K, a.c = c_k some c_k in H /\ K.
so, a = (c_k).c^{-1} in H /\ K ==> contradiction.]
and a.c in H.
so, a.c in A.
and If c = e, a.c = a. so, A.(H/\K) = A.

Similary, (H/\K).B = B.

so, |HK| = |A.B| + |A.(H/\K)| + |(H/\K).B| + |(H/\K)| = |AB| + i + j + t.

While it is more or less clear that A.B and H/\K do not intersect A or
B or each other, you have not proven it.

Consider AB = {(a_n).(b_m) | n = 1,2,...i, m = 1,2,....j}
Find the mutiplicity for each (a_n).(b_m).
Namely,
If (h_1).(k_1) = (h_2).(k_2),
then (h_2)^{-1}.(h_1) = (k_2).(k_1)^{-1} = c_k in H/\K.
so, h_1 = (h_2).(c_k) ==> h_2 = (h_1).(c_k)^{-1} and k_2 = (c_k).(k_1).
so, we can make the (h_2).(k_2) by c_k.(for given h_1, k_1.)
so, the mutiplicity is t.
so, |AB| = (i*j) / t.

In fact, this is what you should use directly: there is a bijection
between the number of times a particular product h*k occurs, and the
number of elements of H/\K.

Given hk, if hk = h'k' then, as you do above, you have (h')^{-1}h =
k'(k)^{-1}, so c = (h')^{-1}h = k'(k)^{-1} lies in H/\K.

So, given any way to obtain hk as a product of two different elements,
you obtain an element of H/\K.

Conversely, say you have hk, and you are given c in H/\K. Then you can
use c to find a to express hk as a product h'k', namely by setting
h' = hc^{-1} and k'=ck. Then h'k' = hk.

Thus, there is a bijection between

{h'k' | h' in H, k' in K, h'k'= hk}

and H/\K.

Thus, |HK|*|H/\K| = |H||K|

This holds even for infinite cardinalities; if |H| or |K| are finite,
then so is |H/\K| and so you get |HK| = |H||K|/|H/\K|, the desired
equality.


Otherwise, you did much too much work but it was fine.

so, |HK| = (i*j)/t + (i+j) + t.

and
|H||K| / |H/\K| = (i+t)(j+t) / t = (i*j)/t + (i+j) + t = |HK|.

(2)
Let H = {h_1, ... , h_n}.

K is a subgroup of HK.

Wait! You don't even know that HK is a group, so it makes no sense to
talk about subgroups of HK.

(Remember that HK is a group if and only if HK = KH as sets; it does
not always hold).


So this proof is wrong.

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

Arturo Magidin
magidin-at-member-ams-org



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