What are the normal subgroups of S3?

The Mathematician — Sat, 05 Nov 2022 13:00:07 +0000

The normal subgroups of S3 are the identity group,

\{e,(123),(132)\}

, and S3 itself. To prove this, a subgroup of

S_3

, say

N

, is a normal subgroup of

G

gNg^{-1} \in N

for all

g \in G

.

Proof. We start with the identity group

\{e\}

. This is straightforward as we get

geg^{-1} = gg^{-1} = e \in \{e\}

for all

g \in G

. Therefore

\{e\}

is a normal subgroup. Next, we will look at the subgroups of order 2, i.e.,

\{e, (12)\}

\{e, (13)\}

and

\{e, (23)\}

. We see that none of them are normal subgroups:

\begin{align*} (13)(12)(31) &= (23) \not \in \{e,(12)\} \\ (23)(13)(32) &= (12) \not \in \{e,(13)\} \\ (12)(23)(21) &= (13) \not \in \{e,(23)\} \\ \end{align*}

Our last subgroup which we need to check is the subgroup of order 3, i.e.,

\{e,(123),(132)\}

. This is a normal subgroup because:

\begin{align*} e(123)e &= (123) \\ (12)(123)(21) &= (132) \\ (13)(123)(31) &= (132) \\ (23)(123)(32) &= (132) \\ (123)(123)(321) &= (123) \\ (132)(123)(231) &= (123) \\ \end{align*}

and

\begin{align*} e(132)e &= (132) \\ (12)(132)(21) &= (123) \\ (13)(132)(31) &= (123) \\ (23)(132)(32) &= (123) \\ (123)(132)(321) &= (132) \\ (132)(132)(231) &= (132) \\ \end{align*}

S_3

is closed under composite, we have that

S_3

is a normal subgroup itself. So we have proved that the identity group, the alternating group of degree 3, and S3 itself are the normal subgroups of S3.

The post What are the normal subgroups of S3? appeared first on Epsilonify.

What are the normal subgroups of S3? Archives - Epsilonify

What are the normal subgroups of S3?