(T/F): $(a \ b \ c),(a' \ b' \ c')\in A_5\Rightarrow \exists \omega\in A_5:(a \ b \ c)=\omega (a' \ b' \ c')\omega^{-1}$

Question

Question:(True/False)

Let $(a \ b \ c),(a' \ b' \ c')\in A_5$.Then $\exists\omega\in A_5:(a \ b \ c)=\omega (a' \ b' \ c')\omega^{-1}$

Attempt:

I think that it is true:

We know that $\langle (a \ b \ c)\rangle,\langle (a' \ b' \ c')\rangle$ are Sylow $3-$subgroups of $A_5$,and that all Sylow $3-$subroups of a group are conjugate,so $\exists \omega_1\in A_5:\langle (a \ b \ c)\rangle=\omega_1 \langle (a' \ b' \ c')\rangle \omega_1^{-1}$

• If $(a \ b \ c)=\omega_1(a' \ b' \ c')\omega_1^{-1}$ we take $\omega=\omega_1 \checkmark$
• If $(a \ b \ c)=\omega_1(a' \ b' \ c')^{2}\omega_1^{-1}=\omega_1(a' \ c' \ b')\omega_1^{-1}=\omega_1(d' \ e')(b' \ c')(a' \ b' \ c')(b' \ c')(d' \ e')\omega_1^{-1}$ so we take $\omega=\omega_1(d' \ e')(b' \ c')\checkmark$

Is this correct?

Thanks in advance!

Answer 1

Yes that's correct. We can also construct $\omega$ explicitly and avoid Sylow theory.

Note the effect of conjugation is given by $\omega(a~b~c)\omega^{-1}=(\omega(a)~\omega(b)~\omega(c))$. Then consider constructing a function $\omega$ with $\omega(a)=a',\omega(b)=b',\omega(c)=c'$. For convenience write

$$ \{1,2,3,4,5\}=\{a,b,c,d,e\}=\{a',b',c',d',e'\} $$

So either $\omega=(\begin{smallmatrix} a & b & c & d & e \\ a' & b' & c' & d' & e'\end{smallmatrix})$ is an even permutation and resolves the problem, or else it is odd and the permutation $(d'~e')\omega=(\begin{smallmatrix} a & b & c & d & e \\ a' & b' & c' & e' & d'\end{smallmatrix})$ is even and resolves the problem.

(Note I am using cycle notation and two-line notation above.)

Answer 2

Let us show it on an example in the natural $5 \times 5$ linear representation of $\frak{S}_5$ with:

$$P_1:=\left(\begin{array}{ccc|cc}0&0&1&0&0\\ 1&0&0&0&0\\ 0&1&0&0&0\\ \hline 0&0&0&1&0\\ 0&0&0&0&1\end{array}\right), \ \ \ \ P_2:=\left(\begin{array}{cc|ccc}1&0&0&0&0\\ 0&1&0&0&0 \\ \hline 0&0&0&0&1\\ 0&0&1&0&0\\ 0&0&0&1&0\end{array}\right).$$

$$\Omega:=\begin{pmatrix}0&0&0&1&0\\ 0&0&0&0&1\\ 1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\end{pmatrix}, \ \ \ \ \Omega':=\begin{pmatrix}0&0&0&0&1\\ 0&0&0&1&0\\ 1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\end{pmatrix}.$$

then we have : $P_2=\Omega P_1 \Omega^{-1}.$

Indeed, $P_1$ represents the 3-cycle on columns $C_1 \to C_2 \to C_3 \to C_1$,

$P_2$ represents the 3-cycle on columns $C_3 \to C_4 \to C_5 \to C_3$,

and $\Omega$ is a transition matrix where $C_1 \to C_3, C_2 \to C_4, C_3 \to C_5$.

It is to be remarked that we also have $P_2=\Omega'P_1\Omega'^{-1}$.

There is no unicity of factor $\Omega$.