I recently posted this question on math.stackexchange.com. The link is this.
My assertion was “Let be a group with three normal subgroups
and
such that
. Then
. This is a generalization of the Third Isomorphism Theorem, which states that
, where
.”
What was my rationale behind asking this question? Let be a group and
its normal subgroup. Then
contains elements of the form
, where
, for every
.
Now let be two normal subgroups of
such that
. Then
contains elements of the form
and
contains elements of the form
. Now consider
. One coset of this would be
all elements of
all elements of
all elements of
all elements of
all elements of
all elements of
. We are effectively adding every element of
to all elements of
. The most important thing to note here is that every element of
is also present in
.
Every element of the form any element in
in
will give the same element in
, and by extension in
. Let
and
be two elements in
(
) such that both are not in
. Then they will not give the same element in
. However, as every element of
is individually added to them in
, they will give the same element in the latter. If
and
form different cosets in
, then they will also form different cosets in
. This led me to conclude that
.
This reasoning is however flawed, mainly because need not be a subgroup of
. Hence, in spite of heavy intuition into the working of cosets, I got stuck on technicalities.