# Conjugate subgroups of $S_4$

$A = \langle (1,2,3),(1,2)\rangle$

$B = \langle (1,2,4),(1,2)\rangle$

$C = \langle (1,3,4),(1,3)\rangle$

$D = \langle (2,3,4),(2,3)\rangle$

I want to proof that these subgroups of $S_4$ ( which are $\cong S_3$)are conjugate to one another but the computations are tedious! :'( These subgroups aren't Sylow 2-supgroups so I can't say they are conjugate to one another....however they're 2-subgroups. Is there some theorem/lemma that says p-subgroups are conjugate? Or is there a smarter way to do this? or am I compelled to use brute force.

Thanks :D

UPDATE:

I computed $A$= $\{(1),(123),(132),(13),(12),(23)\}$

and $B$=$\{(1),(142),(124),(14),(12),(24)\}$

Following @Alex suggestion, $(14)A(14)=B$ Sample computation with $\sigma =(14)$ $(14)(123)(14) = (\sigma(1),\sigma(2),\sigma(3))=(423)=(234)$ $(14)A(14)= \{(1),(234).........\}$ I can see already that this isn't equal to $B$.

So I tried $(14)B(14)$ so

$(14)B(14)=\{(1),(124).....\}$ which again is not $A$

What am I doing wrong?

• Permutations are conjugate if they have the same cycle index. Each of these four groups is generated by a set of conjugate permutations. – MJD Apr 14 '14 at 0:06

Note that they are obtained by relabeling $1,2,3,4$. So we have that A is conjugate to B by $(34)$, B is conjugate to C by $(23)$, and C is conjugate to D by $(12)$.
• @abe15 The key is to know that $$\sigma(i_1\;i_2\;\cdots\;i_n)\sigma^{-1}=(\sigma(i_1)\;\sigma(i_2)\;\cdots\;\sigma(i_n))$$ – Alex Becker Apr 13 '14 at 23:20
• I think you accidentally added a space between "s" and "igma" -_-.... $\sigma$ $\s \ igma$ – abe Apr 13 '14 at 23:52