Subset Product of Subgroups/Examples/Subgroups Generated by b and a b in D3

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Examples of Use of Subset Product of Subgroups

Consider the dihedral group $D_3$, given as the group presentation:

$D_3 = \gen {a, b: a^3 = b^2 = e, a b = b a^{-1} }$


Consider the generated subgroups $H := \gen b$ and $K := \gen {a b}$:

\(\ds \gen b\) \(=\) \(\ds \set {e, b}\) as $b^2 = e$
\(\ds \gen {a b}\) \(=\) \(\ds \set {e, a b}\) as $\paren {a b}^2 = a b b a^{-1} = e$

Then $H$ and $K$ are not permutable, and neither $H K$ nor $K H$ is a subgroup of $D_3$.


Proof

Consider the subset product $H K$:

\(\ds H K\) \(=\) \(\ds \set {h k: h \in H, k \in K}\)
\(\ds \) \(=\) \(\ds \set {e, b, a b, b \paren {a b} }\)
\(\ds \) \(=\) \(\ds \set {e, b, a b, b \paren {b a^{-1} } }\)
\(\ds \) \(=\) \(\ds \set {e, b, a b, a^{-1} }\) as $b^2 = e$
\(\ds \) \(=\) \(\ds \set {e, b, a b, a^2}\)

But $\set {e, b, a b, a^2}$ has $4$ elements.

Thus by Lagrange's Theorem (Group Theory), $H K$ is not a subgroup of $D_3$.


Then we see:

\(\ds K H\) \(=\) \(\ds \set {k h: k \in K, h \in H}\)
\(\ds \) \(=\) \(\ds \set {e, a b, b, \paren {a b} b}\)
\(\ds \) \(=\) \(\ds \set {e, b, a b, a}\) as $b^2 = e$
\(\ds \) \(\ne\) \(\ds H K\)

So $H K \ne K H$ and so $H$ and $K$ are not permutable.

$\blacksquare$


Sources

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