9.2: The Second and Third Isomorphism Theorems

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

Theorem \(\PageIndex{1}\): Second 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 \(G\). Since \(eG∈HN\), \(HN≠∅\). Next, let \(x=h_1n_1,y=h_2n_2∈HN\) (where \(h_1,h_2∈H\) and \(n_1,n_2∈N\)). Then

\(xy^{−1}=h_1n_1(h_2n_2)^{−1}=h_1n_1n^{−1}_2h^{−1}_2.\)

Since \(N≤G\), \(n_1n_2^{−1}\) is in \(N\); so \(h_1n_1n^{−1}_2h^{−1}_2∈h_1Nh^{−1}_2\), which equals \(h_1h_2^{−1}N\), since \(N⊴G\) implies \(Nh^{−1}_2=h^{−1}_2N\). So \(xy^{−1}=∈h_1h_2^{−1}N\). But \(H≤G\) implies \(h_1h_2^{−1}∈H\); thus, \(xy^{−1}∈HN\), and so \(HN\) is a subgroup of \(G\).

Since \(N⊴G\) and \(N⊆HN\), \(N\) is normal in \(HN\) (do you see why?). So \(HN/N\) is a group under left coset multiplication. We define \(ϕ:H→HN/N\) by \(ϕ(h)=hN\) (notice that when \(h∈H\), \(h∈HN\) since \(h=he_G\)). Clearly, \(ϕ\) is a homomorphism. Further, \(ϕ\) is onto: Indeed, let \(y∈HN/N\). Then \(y=h_nN\) for some \(h∈H\) and \(n∈N\). But \(nN=N\), so \(y=hN=ϕ(h)\). Finally,

\(\begin{array} &\text{Ker}ϕ&=\{h∈H:ϕ(h)=N\}\\&=\{h∈H:hN=N\}\\&=\{h∈H:h∈N\}\\&=H∩N \end{array}\).

Thus,

\(H/(H∩N)≃HN/N\),

by the First Isomorphism Theorem.

Theorem \(\PageIndex{2}\): Third 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 \(ϕ:G/K→G/N\) by \(ϕ(aK)=aN\). We have that \(ϕ\) is well-defined: indeed, let \(aK=bK∈G/K\). Then by Statement 6 of Theorem \(7.2.3\), we have \(aN=bN\), that is, \(ϕ(aK)=ϕ(bK)\). So \(ϕ\) is well-defined. \(ϕ\) is clearly onto, and we have

\(\begin{array} & \text{Ker}ϕ&=\{aK∈G/K:ϕ(aK)=N\}\\&=\{aK∈G/K:aN=N\}\\&=\{aK∈G/K:a∈N\}\\&=N/K. \end{array}\)

So the desired results hold, by the First Isomorphism Theorem.

Example \(\PageIndex{1}\)

Using the Third Isomorphism Theorem, we see that the group

\begin{equation*} (\mathbb{Z}/12\mathbb{Z})/(6\mathbb{Z}/12\mathbb{Z}) \end{equation*}

is isomorphic to the group \(\mathbb{Z}/6\mathbb{Z}\text{,}\) or \(\mathbb{Z}_6\text{.}\)