MTH3003 Lecture 8

Following on from last lecture, we can use our knowledge of Lagrange’s theorem to prove a few more things about finite groups, and then apply it to the dihedral groups we looked at previously.

Consequences of Lagrange’s Theorem

We know from Lagrange’s Theorem that if $H\le G$ for a finite group $G$, then $|G|=[G:H]\cdot |H|$. We can easily deduce two corollaries from this:

1. The index of $H$ in $G$ is given by $[G:H]=\frac{|G|}{|H|}$.
2. For any element $g\in G$, its order $o(g)$ must divide $|G|$.

This is incredibly useful for disproving the existence of certain elements or subgroups within finite groups. For example, if we look at the dihedral group $D_{10}$ (which has order 10):

• It cannot contain an element of order $7$, because $7$ does not divide $10$.
• It cannot contain a subgroup of order $7$, for the exact same reason.

Warning

The converse statement to Lagrange’s Theorem does not hold! Just because a number divides $|G|$, this does not guarantee that $G$ has a subgroup of that order. The smallest counterexample is the Alternating Group of degree 4 (order 12), which has no subgroup of order 6.

Application: Understanding Dihedral Groups

Recall that the dihedral group $D_{2n}=\{e,\rho ,\dots ,{\rho }^{n-1},{\sigma }_1,\dots ,{\sigma }_n\}$, where $$\begin{gathered} \rho =(1 \\ ,\dots n) \end{gathered}$$ is a rotation and the ${\sigma }_i$ are reflections. We can use Lagrange’s Theorem to simplify how we represent these reflections.

Let $C=⟨\rho ⟩=\{e,\rho ,\dots ,{\rho }^{n-1}\}$. This is a subgroup of $D_{2n}$ with size $|C|=n$. Because $|D_{2n}|=2n$, we can calculate the index of $C$:

$$[D_{2n}:C]=\frac{|D_{2n}|}{|C|}=\frac{2n}{n}=2$$

This means there are exactly two cosets of $C$ in $D_{2n}$. Let $\sigma ={\sigma }_1$. Because a reflection cannot be a rotation, $\sigma \notin C$. Therefore, the two cosets must be $C$ and $\sigma C$, giving us:

$$D_{2n}=C\cup \sigma C$$

Because all the reflections are not in $C$, they must all lie in $\sigma C$. This means every single reflection can be written in the form $\sigma {\rho }^i$. Thus, we can rewrite the entire group using only two generators:

$$D_{2n}=\{e,\rho ,{\rho }^2,\dots ,{\rho }^{n-1},\sigma ,\sigma \rho ,\sigma {\rho }^2,\dots ,\sigma {\rho }^{n-1}\}$$

Combining Rotations and Reflections

This simplified list is great, but calculating products like $\rho \sigma$ requires a trick. Because $\rho \sigma$ reverses the order of corners, it must be a reflection, meaning it has order 2. Therefore, $(\rho \sigma )(\rho \sigma )=e$, which implies:

$$\rho \sigma =(\rho \sigma {)}^{-1}={\sigma }^{-1}{\rho }^{-1}=\sigma {\rho }^{-1}=\sigma {\rho }^{n-1}$$

Using the rule $\rho \sigma =\sigma {\rho }^{-1}$, along with ${\rho }^n=e$ and ${\sigma }^2=e$, we can simplify any combination of rotations and reflections into the standard form.

Simplifying ${\rho }^3\sigma$

We can move $\sigma$ to the left by pulling it through the $\rho$s, inverting them as we go: ${\rho }^3\sigma ={\rho }^2\sigma {\rho }^{-1}=\rho \sigma {\rho }^{-2}=\sigma {\rho }^{-3}=\sigma {\rho }^{n-3}$

In cycle notation, if $\sigma$ is the line of reflection through corner 1, it looks like $\sigma =(2n)(3n-1)(4n-2)\dots$

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Pre-Lecture Notes from University Notes

• Corollaries of Lagrange’s Theorem: $[G:H]=|G|/|H|$ and $o(g)$ divides $|G|$.
• Warning that the converse of Lagrange’s Theorem is false (Alternating Group of degree 4 is a counterexample).
• Application of Lagrange’s Theorem to show that $D_{2n}$ can be partitioned into two cosets: rotations $C$ and reflections $\sigma C$.
• Redefining $D_{2n}$ using only generators $\rho$ and $\sigma$: $\{e,\rho ,\dots ,{\rho }^{n-1},\sigma ,\sigma \rho ,\dots ,\sigma {\rho }^{n-1}\}$.
• Calculating products using the core relation $\rho \sigma =\sigma {\rho }^{-1}$ to simplify combinations of rotations and reflections.