What are the normal subgroups of D3?

The normal subgroups of

D_3

are

\{e\}

and

\{e,r,r^2\}

where

\begin{equation*} D_3 = \langle r,s \ | \ r^3 = s^2 = e, rs = sr^{-1} \rangle \end{equation*}

We will prove why that is the only normal subgroup. To do that, the subgroup

N

G

is a normal subgroup if

gNg^{-1} \subseteq N

for all

g\in G

.

Proof. We have that

\{e\}

\{e,r,r^2\}

\{e,s\}

\{e,rs\}

and

\{e,r^2s\}

are the subgroups of

D_3

. We know that

geg^{-1}

is in all the subgroups for all

g \in G

. So now, we need to check the other values of the subgroups. It is easy to see that

rsr^{-1} = r^2s \not \in \{e,s\}

r(rs)r^{-1} = s \not \in \{e,rs\}

and

r^2(rs)r^{-2} \not \in \{e,r^2s\}

. So those order 2 subgroups are not normal subgroups. What is left to check is

\{e,r,r^2\}

, which is indeed a normal subgroup of

D_3

\begin{align*} ere &= r \in \{e,r,r^2\} \\ rrr^{-1} &= r \in \{e,r,r^2\} \\ r^2rr^{-2} &= r^{2 + 1 - 2} = r \in \{e,r,r^2\} \\ srs^{-1} &= srs = ssr^{-1} = r^{2} \in \{e,r,r^2\} \\ rsr(rs)^{-1} &= rsrs^{-1}r^{-1} = sr^{-1}rs^{-1}r^{-1} = r^{-1} = r^{2} \in \{e,r,r^2\} \\ r^2sr(r^2s)^{-1} &= rsr^{-1}rs^{-1}r^{-2} = r^{-1} = r^{2} \in \{e,r,r^2\} \\ \end{align*}

\begin{align*} er^2e &= r^2 \in \{e,r,r^2\} \\ rr^2r^{-1} &= r^2 \in \{e,r,r^2\} \\ r^2r^2r^{-2} &= r^{2 + 2 - 2} = r^2 \in \{e,r,r^2\} \\ sr^2s^{-1} &= sr^{-1}s^{-1} = rss^{-1} = r \in \{e,r,r^2\} \\ rsr^2(rs)^{-1} &= rsr^2s^{-1}r^{-1} = rsr^2sr^{-1} = rsr^2rs = rss = r \in \{e,r,r^2\} \\ r^2sr^2(r^2s)^{-1} &= r^2sr^{-1}s^{-1}r^{-2} = r^2rss^{-1}r^{-2} = r^{-2} = r \in \{e,r,r^2\}\\ \end{align*}

So we get that the trivial group and group of order 3 are the normal subgroups of D3.