The third isomorphism theorem

In Section 8.3, we considered (Theorems 8.3.4 and 8.3.7) the first and second isomorphism theorems. We come now to what is frequently called the third isomorphism theorem, the proof of which is due to Zassenhaus.

Theorem 11.1.1 (Third Isomorphism Theorem)   Let $G$ be a group with subgroups $G_1$, $G_2$, $H_1$, and $H_2$. Let $H_1 \lhd G_1$ and $H_2\lhd G_2$. Then $( G_1 \cap H_2)H_1 \lhd (G_1 \cap G_2)H_1$ and $(G_2 \cap H_1)H_2 \lhd (G_1 \cap G_2)H_2$. Moreover,

\begin{displaymath} (G_1 \cap G_2)H_1/( G_1 \cap H_2)H_1 \cong (G_1 \cap G_2)H_2/(G_2 \cap H_1)H_2 . \end{displaymath}

Proof: We first note that $G_1 \cap G_2$, $G_1 \cap H_2$, and $H_1$ are all subgroups of $G_1$. Also by hypothesis, $H_1 \lhd G_1$. Thus by Proposition 8.1.3, $(G_1 \cap G_2)H_1$ and $(G_1 \cap H_2)H_1$ are subgroups of $G_1$. We next claim that

\begin{displaymath} (G_1 \cap H_2)H_1 \lhd (G_1 \cap G_2)H_1. \end{displaymath}

To see this let $a \in G_1 \cap G_2$, $b \in G_1 \cap H_2$, and $c,d \in H_1$. Then $aba^{-1}\in G_1 \cap H_2$, since $a,b \in G_1$. But now $aba^{-1} \in G_1$ and $b \in H_2$ and $a \in G_2$, which implies that $aba^{-1} \in H_2$ since $H_2\lhd G_2$. Also $aca^{-1} \in H_1$, since $a\in G_1$, $c \in H_1$, and $H_1 \lhd G_1$. Since a typical element of $(G_1 \cap H_2)H_1$ is of the form $bc$, where $b \in G_1 \cap H_2$ and $c \in H_1$, we therefore get that
\begin{displaymath} a(G_1 \cap H_2)H_1a^{-1} \subset (G_1 \cap H_2)H_1 \end{displaymath} (11.1)

where $a \in G_1 \cap G_2$. Moreover, using the same notation as above,

\begin{displaymath} dbd^{-1} = d(bd^{-1}b^{-1})b. \end{displaymath}

Bbut $bd^{-1}b^{-1}\in H_1$, since $d \in H_1$, $b \in G_1$, and $H_1 \lhd G_1$, so

\begin{displaymath} dbd^{-1} \in H_1(G_1 \cap H_2). \end{displaymath}

This implies
\begin{displaymath} d(G_1 \cap H_2)H_1d^{-1} =d(G_1 \cap H_2)d^{-1}dH_1d^{-1}, \end{displaymath} (11.2)

where we have used the fact that $H_1(G_1 \cap H_2) = (G_1 \cap H_2)H_1$ (which is true since $H_1 \lhd G_1$, by Proposition 8.3.6). Recall the typical element of $(G_1 \cap G_2)H_1$ is of the form $ad$ where $a \in G_1 \cap G_2$ and $d \in H_1$. According to Proposition 6.1.4, this, (11.1) and ((11.2) together imply $( G_1 \cap H_2)H_1 \lhd (G_1 \cap G_2)H_1$. This proves the claim above.

On the basis of the second isomorphism (Theorem 8.3.7), we have


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(note that $(G_1 \cap H_2)H_1\lhd (G_1 \cap G_2)(G_1 \cap H_2)H_2$ since $( G_1 \cap H_2)H_1 \lhd (G_1 \cap G_2)H_1$ and $(G_1 \cap G_2) (G_1 \cap H_2)H_1 =(G_1 \cap G_2)H_1 $. Now since $(G_1 \cap H_2)H_1 \lhd (G_1 \cap G_2)(G_1 \cap H_2)H_1$, $(G_1 \cap G_2)\cap (G_1 \cap H_2)H_1 \lhd G_1 \cap G_2$, and

\begin{displaymath} (G_1 \cap G_2)/((G_1 \cap G_2)\cap (G_1 \cap H_2)H_1) \cong ((G_1 \cap G_2)H_1)/((G_1 \cap H_2)H_1). \end{displaymath} (11.3)

However, we contend that
\begin{displaymath} (G_1 \cap G_2) (G_1 \cap H_2)H_1 = (G_1 \cap H_2)H_1\cap G_2. \end{displaymath} (11.4)

For clearly

\begin{displaymath} (G_1 \cap G_2) (G_1 \cap H_2)H_1 \subset (G_1 \cap H_2)H_1\cap G_2. \end{displaymath}

While if $w \in (G_1 \cap H_2)H_1\cap G_2$, then $w \in G_2$ and $w = xy$, where $x \in G_1\cap H_2$ and $y \in H_1 \subset G_1$. But then $x \in G_1$ and $y \in G_1$, so $w \in G_1 \cap G_2$, and we then get the inclusion the other way. This proves (11.4).

Next we note that any element of $(G_1 \cap H_2)H_1$ is of the form $uv$, where $u \in G_1 \cap H_2$ and $v \in H_1$. If this element also belongs to $G_2$, i.e., $uv \in G_2$, then

\begin{displaymath} v = u^{-1}(uv) \in G_2; \end{displaymath}

hence $v \in G_2 \cap H_1$. Thus $uv \in (G_1 \cap H_2)(G_2 \cap H_1)$. We have shown that

\begin{displaymath} (G_1 \cap H_2)H_1 \cap G_2 \subset (G_1 \cap H_2)(G_2 \cap H_1) \end{displaymath}

Since the reverse inclusion is clear, we have

\begin{displaymath} (G_1 \cap H_2)H_1 \cap G_2 = (G_1 \cap H_2)(G_2 \cap H_1) \end{displaymath}

Combining this last result with (11.4) and (11.3) yields
\begin{displaymath} (G_1 \cap G_2)/((H_1 \cap G_2) (G_1 \cap H_2)) \cong ((G_1 \cap G_2)H_1)/((G_1 \cap H_2)H_1). \end{displaymath} (11.5)

However, by symmetry, i.e., replacing 1 by 2 and vice versa in (11.5) yields
\begin{displaymath} (G_1 \cap G_2)/((H_1 \cap G_2) (G_1 \cap H_2)) \cong ((G_1 \cap G_2)H_2)/((G_2 \cap H_1)H_2). \end{displaymath} (11.6)

and by (11.5) and (11.6), we finally have that

\begin{displaymath} (G_1 \cap G_2)H_1/( G_1 \cap H_2)H_1 \cong (G_1 \cap G_2)H_2/(G_2 \cap H_1)H_2 . \end{displaymath}

$\Box$

We note that the symmetry arguement used in the above proof to get (11.6) could, of course, have been replaced by an argument similar to that given in the first part of the proof. (See exercise 1 for this section.)


Subsections
David Joyner 2007-08-06