The Third Group Isomorphism Theorem

The Third Group Isomorphism Theorem

Recall from The Second Group Isomorphism Theorem page that if $G$ is a group and $H$ and $N$ are subgroups of $G$ such that $N \trianglelefteq G$ then:

  • a) $H \cap N \trianglelefteq H$.
  • b) $H / (H \cap N) \cong (HN)/N$.

We will now state and prove the third group isomorphism theorem.

Theorem 1 (The Third Group Isomorphism Theorem): Let $G$ be a group and let $H$, $K$, $H'$, and $K'$ be subgroups of $G$ such that:
1) $H' \trianglelefteq H$.
2) $K' \trianglelefteq K$.
Then:
a) $H'(H \cap K') \trianglelefteq H'(H \cap K)$.
b) $K'(H' \cap K) \trianglelefteq K'(H \cap K)$.
c) $H'(H \cap K) / H'(H \cap K') \cong K'(H \cap K) / K'(H' \cap K)$.
Screen%20Shot%202017-01-29%20at%207.09.45%20PM.png

Refer to the diagram above when reading the proof of the third group isomorphism theorem.

The third isomorphism theorem has many names. The same result goes by the "Butterfly Lemma" and the "Zassenhaus Lemma".

  • Proof of a) Since $H \cap K$ and $K'$ are subgroups of $K$ and $K' \trianglelefteq K$, we have by the second group isomorphism theorem that:
(1)
\begin{align} \quad (H \cap K) \cap K' \trianglelefteq (H \cap K) \quad \Leftrightarrow \quad (H \cap K') \trianglelefteq (H \cap K) \quad (*) \end{align}
(2)
\begin{align} \quad (H \cap K) / [(H \cap K) \cap K'] \cong [(H \cap K)K']/K' \quad \Leftrightarrow \quad (H \cap K)/(H \cap K') \cong [(H \cap K)K']/K' \quad (**) \end{align}
  • From $(*)$ we have that $H'(H \cap K') \trianglelefteq H'(H \cap K)$ which proves (a). $\blacksquare$
  • Proof of b) Since $H \cap K$ and $H'$ are subgroups of $H$ and $H' \trianglelefteq H$, we have by the second group isomorphism theorem that:
(3)
\begin{align} \quad (H \cap K) \cap H' \trianglelefteq (H \cap K) \quad \Leftrightarrow \quad (H' \cap K) \trianglelefteq (H \cap K) \quad (***) \end{align}
(4)
\begin{align} \quad (H \cap K)/[(H \cap K) \cap H'] \cong [(H \cap K)H']/H' \quad \Leftrightarrow \quad (H \cap K)/(H' \cap K) \cong [(H \cap K)H']/H' \quad (****) \end{align}
  • From $(***)$ we have that $K'(H' \cap K) \trianglelefteq K'(H \cap K)$ which proves (b). $\blacksquare$
  • Proof of c) From $(*)$ we have that $H \cap K' \trianglelefteq H \cap K$. Therefore:
(5)
\begin{align} \quad (H \cap K')(H' \cap K) \trianglelefteq (H \cap K) \end{align}
  • And so:
(6)
\begin{align} \quad K'(H' \cap K) \trianglelefteq K'(H \cap K) \end{align}
  • Hence:
(7)
\begin{align} \quad K'(H \cap K) / K'(H' \cap K) \cong (H \cap K)/[(H' \cap K)(H \cap K') \quad (\dagger) \end{align}
  • From $(***)$ we have that $(H' \cap K) \trianglelefteq (H \cap K)$. Therefore:
(8)
\begin{align} \quad (H \cap K')(H' \cap K) \trianglelefteq (H \cap K) \end{align}
  • And so:
(9)
\begin{align} H'(H \cap K') \trianglelefteq H'(H \cap K) \end{align}
  • Hence:
(10)
\begin{align} \quad H'(H \cap K) / H'(H \cap K') \cong (H \cap K)/[(H' \cap K)(H \cap K') (\dagger \dagger) \end{align}
  • From $(\dagger)$ and $(\dagger \dagger)$ we see that:
(11)
\begin{align} \quad H'(H \cap K) / H'(H \cap K') \cong K'(H \cap K) / K'(H' \cap K) \end{align}
  • Which proves (c). $\blacksquare$
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