0

Find the number of normal subgroups of $A_4$. Is there any way to find this without actually finding the subgroups?

In78
  • 879
  • 1
  • 11
  • 25

1 Answers1

4

Consider the subgroup $K= \{id, (12)(34), (13)(24),(14)(23)\} \leq A_4$.

Writing the elements of $S_4$ as product of disjoint cycles we have that $A_4= \{3 - \text{cycles}\} \cup K$. As long as K is the only subgroup of order $4$, then $K \lhd A_4$.

Now let $H\neq \{id\} $ be a normal subgroup of $A_4$.

  • If $H$ has a $3$-cycle, say $(123)$, then $H$ has its inverse $(132)$ thefore it also has $(124) = (324)(132)(324)^{-1}$, Thus $H \supseteq \langle (123),(124) \rangle = A_4$ (why?). Conclusion $H = A_4$.

  • If $H$ has no $3$-cycle then is contains an element of $K$ different from the identity, say $(12)(34)$. Then it has $(13)(24) = (124)(12)(34)(124)^{-1}$ and also $(14)(23) = (12)(34)(13)(24)$. Therefore $H = K$.

Aaron Maroja
  • 17,869