Tower Law for Subgroups/Proof 2

Theorem

Let $\struct {G, \circ}$ be a group.

Let $H$ be a subgroup of $G$ with finite index.

Let $K$ be a subgroup of $H$.

Then:

$\index G K = \index G H \index H K$

where $\index G H$ denotes the index of $H$ in $G$.

Proof

Assume $G$ is finite.

Then:

\(\ds \index G H\)

\(=\)

\(\ds \frac {\order G} {\order H}\)

Lagrange's Theorem

\(\ds \index G K\)

\(=\)

\(\ds \frac {\order G} {\order K}\)

Lagrange's Theorem

\(\ds \leadsto \ \ \)

\(\ds \index G K\)

\(=\)

\(\ds \frac {\order H} {\order K} \times \index G H\)

Since $K \le H$, from Lagrange's Theorem we have that $\dfrac {\order H} {\order K} = \index H K$.

Hence the result.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 43$. Lagrange's theorem: Worked Example $2$
• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $5$: Cosets and Lagrange's Theorem: Corollary $5.13$