#
Section9.2The Second and Third Isomorphism Theorems

The following theorems can be proven using the First Isomorphism Theorem. They are very useful in special cases.

#####
Theorem9.2.1Second Isomorphism Theorem

Let \(G\) be a group, let \(H\leq G\text{,}\) and let \(N\unlhd G\text{.}\) Then the set

\begin{equation*}
HN=\{hn:h\in H, n\in N\}
\end{equation*}
is a subgroup of \(G\text{,}\) \(H\cap N\unlhd H\text{,}\) and

\begin{equation*}
H/(H\cap N) \simeq
HN/N.
\end{equation*}
##### Proof

We first prove that \(HN\) is a subgroup of \(G\text{.}\) Since \(e_G\in
HN\text{,}\) \(HN\neq \emptyset\text{.}\) Next, let \(x=h_1n_1, y=h_2n_2\in HN\) (where \(h_1,h_2\in H\) and \(n_1,n_2\in N\)). Then

\begin{equation*}
xy^{-1}=h_1n_1(h_2n_2)^{-1}=h_1n_1n_2^{-1}h_2^{-1}.
\end{equation*}
Since \(N\leq G\text{,}\) \(n_1n_2^{-1}\) is in \(N\text{;}\) so \(h_1n_1n_2^{-1}h_2^{-1}\in
h_1Nh_2^{-1}\text{,}\) which equals \(h_1h_2^{-1}N\text{,}\) since \(N\unlhd G\) implies \(Nh_2^{-1}=h_2^{-1}N\text{.}\) So \(xy^{-1}=\in h_1h_2^{-1}N.\) But \(H\leq G\) implies \(h_1h_2^{-1}\in H\text{;}\) thus, \(xy^{-1}\in HN\text{,}\) and so \(HN\) is a subgroup of \(G\text{.}\)

Since \(N\unlhd G\) and \(N\subseteq HN\text{,}\) \(N\) is normal in \(HN\) (do you see why?). So \(HN/N\) is a group under left coset multiplication. We define \(\phi: H\to HN/N\) by \(\phi(h)=hN\) (notice that when \(h\in H\text{,}\) \(h\in HN\) since \(h=he_G\)). Clearly, \(\phi\) is a homomorphism. Further, \(\phi\) is onto: Indeed, let \(y\in HN/N\text{.}\) Then \(y=hnN\) for some \(h\in H\) and \(n\in N\text{.}\) But \(nN=N\text{,}\) so \(y=hN=\phi(h)\text{.}\) Finally,

\begin{align*}
\Ker \phi\amp =\{h\in H\,:\, \phi(h)=N\}\\
\amp =\{h\in H\,:\, hN=N\}\\
\amp =\{h\in
H\,:\, h\in N\}\\
\amp =H\cap N.
\end{align*}
Thus,

\begin{equation*}
H/(H\cap N)
\simeq HN/N,
\end{equation*}
by the First Isomorphism Theorem.

#####
Theorem9.2.2Third Isomorphism Theorem

Let \(G\) be a group, and let \(K\) and \(N\) be normal subgroups of \(G\text{,}\) with \(K\subseteq N\text{.}\) Then \(N/K \unlhd G/K\text{,}\) and

\begin{equation*}
(G/K)/(N/K)\simeq G/N.
\end{equation*}
##### Proof

Define \(\phi: G/K\to G/N\) by \(\phi(aK)=aN\text{.}\) We have that \(\phi\) is well-defined: indeed, let \(aK=bK \in G/K\text{.}\) Then by Statement 6 of TheoremĀ 7.2.7, we have \(aN=bN\text{,}\) that is, \(\phi(aK)=\phi(bK)\text{.}\) So \(\phi\) is well-defined. \(\phi\) is clearly onto, and we have

\begin{align*}
\Ker \phi\amp =\{aK\in
G/K\,:\,\phi(aK)=N\}\\
\amp =\{aK\in G/K\,:\,aN=N\}\\
\amp =\{aK\in
G/K\,:\,a\in N\}\\
\amp =N/K.
\end{align*}
So the desired results hold, by the First Isomorphism Theorem.

#####
Example9.2.3

Using the Third Isomorphism Theorem, we see that the group

\begin{equation*}
(\Z/12\Z)/(6\Z/12\Z)
\end{equation*}
is isomorphic to the group \(\Z/6\Z\text{,}\) or \(\Z_6\text{.}\)