1.7. Quotient Groups.
The work done in the previous section on Normal subgroups now
leads to the formulation of the Quotient Group. Up to
this point we've studied groups which have familiar, concerete
objects, but now we're going to get a little bit abstract.
We're going to look at the useful concept of the quotient group, ,
which is a
group whose elements are cosets. That is, the elements of
our group are going to be sets themselves. The operation on the
elements of the quotient group can only make sense if the cosets
are from a subgroup which is normal to .
Let be a group and . Define to be the
set consisting of all the possible right (or equivalently
left) cosets. If we
equip this set with a product such that
then forms a group, called the Quotient Group.
Let's review what this is saying. Basically, if we have a normal
subgroup of , the set of cosets
with the product forms a group.
- Identity. To show that this set is a group, we first define the identity
element to simply be . This is a "trivial" coset, and for
any , where ,
so is a natural and apporopriate choice for an identity as
it has the property of an identity element.
* Associativity. Associativity is derived from the
associativity of our group itself. Observe that for any
we have
Therefore for all , so the product relation is associative.
* Closedness. The result of our proposed product is
always a coset itself (), and since
is a set of all cosets we see that this set is
closed under .
* Inverses. For any , where , we
see that the inverse element is , since
and we already defined to be our identity element. So
our proposed inverse makes sense.
Note that
since , so an inverse
element not only exists but it also exists in
All together, this allows us to observe that we have a group
structure, so long as .
{\color{purple}(Why do we need this
the condition that ? Well, because the only way we can make damn sure
that is by Theorem 1.10, which requires
that .)
}
{\color{NavyBlue} Note that there is another way to think about . The elements
of the quotient group are cosets, right? However, let us not forget
that cosets are simply equivalence classes which
respect the following equivalence relation }: {\color{Black} if is a group, is a
subgroup, then for any we say that if and only if .} {\color{NavyBlue} Thus we can
recast our definition follows:}
\
\begingroup
\par
\leftskip25pt
\rightskip\leftskip
\noindent Let . Then the set is defined to consist of all
of the
\sout{right (or left) cosets of in } equivalence classes of
the elements of (under the equivalence relation stated in the
previous paragraph).
\par
\endgroup
\vspace{1cm}
{\color{Violet}We thus have two equivalent ways to interpret the meaning of a
quotient group. One involves equivalence classes, while the other
involves cosets. In our case it seems more complicated to think
about equivalence classes.
However, in different applications of group theory (such
as to algebraic geometry and topology) it will be convenient to
interpret quotient groups as equivalence classes. For now, we'll
stick with the coset interpretation, since it's the easiest way to
understand a quotient group.
}
\
Example. Recall that we showed . Thus the quotient group
makes sense by Theorem 1.11,
so let's see what this group looks like.
First, the identity element of our group is .
\
\
\indent In dealing with quotient groups, you may be wondering the
following questions:\
\textcolor{ForestGreen}{Q: If is a normal subgroup of
, and is abelian, is abelian? If is abelian, is
abelian?}
\
A: The answer to the first question is yes.
Observe that
by definition, But since
is normal, we know that for all .
Thus observe that for , we have that
Thus the set must be abelian.
\
\
The answer to the second question is \textbf{no, not always}}. If is abelian,
we know that
for all . However, this only guarantees \textbf{set equality},
not a term-by-term equality (in which case the group would be abelian).
An example of this is \(D_{6**\) with the subgroup
In this case because all the left cosets are and therefore
(Hence by the previous proposition). In addition,
, , so is abelian, but the set
is itself not an abelian group. Thus, it is possible for
to be ableian while itself is not abelian **
\
\
Another fun example for when the quotient group is abelian,
even though the group is abelian, is the following.
\
\
Example.
Let
is subset of and is a subgroup of .
- . First we'll show that is normal
to . Thus let , so that
$
x =
$
for some where . Now let so that
$
h =
$
for some . Then observe that
Therefore, we have that for all , which
implies that is a normal subgroup of .
* is abelian. Now we'll show that is an
abelian group. Firstly, what does it mean for a quotient group to
abelian? Well, it would mean that for any we have
that
Or, in other words,
Thus we need some kind of set equality to be happening. Thus
consider , where again , and suppose and where and . Then observe that
\begin{minipage}{0.40\textwidth}
\end{minipage}
\hfill
\begin{minipage}{0.5\textwidth}
\end{minipage}
\textcolor{purple}{Note that the (1,1) entry in both matrices are
equal; that is, since they are members of
.}
Therefore, we see that
Since are arbitrary members of , we can
replace their sums with another arbitrary .
Then we see that
After cleaning up the sets, we can now see they are equal, which
wasn't as obvious as it was before. They're equal because their
criteria for set memberships are identical; they just have
different variables, but that of course does not change their
members. Therefore we see that for all ,
which proves that is an abelian group, even though nor
are abelian.