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Section Quotient Groups
Worksheet Part 1: Normal Subgroup Practice
The goal of these problems is to familiarize you with the definition of normal subgroup, the Normal Subgroup Test, and find more examples of normal subgroups.
1. Show that the subgroup
\(\SL(2,\R)\) of
\(\GL(2,\R)\) is normal in
\(\GL(2,\R)\text{.}\)
Solution .
Let \(X\in \GL(2,\R), A\in \SL(2,\R)\text{.}\) Then
\begin{equation*}
\det(XAX^{-1})=\det(X)\det(A)\det(X^{-1})=\det(X)\cdot 1 \cdot \frac{1}{\det(X)}=1
\end{equation*}
so \(XAX^{-1} \in \SL(2,\R)\text{.}\) Since this is true for all \(X\in \GL(2,\R)\) and \(A\in \SL(2,\R)\text{,}\) we have that \(\SL(2,\R)\) is normal in \(\GL(2,\R)\text{.}\)
2. Show that the center of a group
\(G\) is normal in the group.
Solution .
Let \(z\in Z(G)\) and \(g\in G\text{.}\) Then
\begin{equation*}
gzg^{-1}=zgg^{-1}=z
\end{equation*}
so \(gzg^{-1} \in Z(G)\text{.}\) Since this is true for all \(z\in Z(G)\) and \(g\in G\text{,}\) we have that \(Z(G)\) is normal in \(G\text{.}\)
3. Suppose that
\(G\) has a unique subgroup
\(H\) of order
\(n\text{.}\) Show that
\(H\) is normal in
\(G\text{.}\)
Solution .
Suppose that
\(H\) is the unique subgroup of order
\(n\) in
\(G\text{.}\) Then for any
\(g\in G\text{,}\) the subgroup
\(gHg^{-1}\) has order
\(n\text{,}\) so by uniqueness we must have that
\(gHg^{-1}=H\text{.}\) Since this is true for all
\(g\in G\text{,}\) we have that
\(H\) is normal in
\(G\text{.}\)
4. Suppose that the index of a subgroup
\(H\) in a group
\(G\) is
\(2\text{.}\) Show that
\(H\) is normal in
\(G\text{.}\)
Hint .
Consider separately the cases
\(g\in H\) and
\(g \notin H\text{.}\) In the second case, you should consider proof by contradiction.
Solution .
Let
\(G\) be a group and
\(H\) a subgroup of index 2. Then there are exactly two left cosets of
\(H\) in
\(G\text{,}\) say
\(H\) and
\(gH\) for some
\(g\in G\text{.}\) Since the right cosets must also partition
\(G\text{,}\) we have that the right coset
\(Hg\) is either equal to
\(H\) or to
\(gH\text{.}\) If
\(Hg=H\text{,}\) then
\(g\in H\text{,}\) which contradicts our choice of
\(g\text{.}\) Therefore,
\(Hg=gH\text{,}\) which means that
\(H\) is normal in
\(G\text{.}\)
Worksheet Part 2: Quotient Practice
The goal of this problem is to practice working with cosets, quotients, and learn a theorem about quotients.
1.
Let
\(G = \Z_4 \oplus U(4)\) and consider the subgroups
\(H=\langle (2,3)\rangle\) and
\(K=\langle (2,1)\rangle\text{.}\)
(a) Show that
\(H\cong K\text{.}\)
Solution .
Both
\(H\) and
\(K\) are cyclic of order 2, so they are isomorphic to each other and to
\(\Z_2\text{.}\)
(b) Compute the quotient
\(G/H\text{.}\) (Why is
\(H\) normal in
\(G\text{?}\) )
Solution .
All subgroups of an Abelian group are normal (by the same argument as for the center of a group being normal in a group). Now the elements of \(G/H\) are
\begin{align*}
(0,1)+H\amp=\{(0,1),(2,3)\}, \quad (1,1)+H=\{(1,1),(3,3)\},\\
(2,1)+H \amp=\{(2,1),(0,3)\}, \quad (3,1)+H=\{(3,1),(1,3)\} \text{.}
\end{align*}
We can see that \(G/H\) has order 4, and the element \((1,1)+H\) has order 4, so \(G/H\cong \Z_4\text{.}\)
(c) Compute the quotient
\(G/K\text{.}\) (Why is
\(K\) normal in
\(G\text{?}\) )
Solution .
Here the elements of \(G/K\) are
\begin{align*}
(0,1)+K \amp =\{ (0,1),(2,1)\},\quad (1,1)+K=\{(1,1),(3,1)\} \\
(0,3)+K \amp =\{ (0,3),(2,3)\}, \quad (1,3)+K=\{(1,3),(3,3)\} \text{.}
\end{align*}
Here we have a group with order 4, but all non-identity elements have order 2, so \(G/K\cong \Z_2\times \Z_2\text{.}\)
(d) Show that
\(G/H \not\cong G/K\text{.}\)
Solution .
Because one group is cyclic while the other is not, they cannot be isomorphic.
(e)
True/False: If
\(G\) is a group with normal subgroups
\(H\) and
\(K\) and
\(H\cong K\text{,}\) then
\(G/H \cong G/K\text{.}\)
Solution .
This computation shows that this statement is false. The subgroups
\(H\) and
\(K\) are isomorphic, but the quotients
\(G/H\) and
\(G/K\) are not isomorphic.
