Prove exists embedding $\phi: \mathbb {Z}_3 \times \mathbb {Z}_3 \times \mathbb {Z}_3 \to S_{27}$ s.t. ${\rm Im}\phi$ intersects 2 conjugacy classes.

Question

Question:

Prove exists an embedding $\phi: \mathbb {Z}_3 \times \mathbb {Z}_3 \times \mathbb {Z}_3 \to S_{27}$ such that ${\rm Im}\phi$ intersects exactly 2 conjugacy classes of $S_{27}$.

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I'm pretty stuck on formalizing and continuing my proof. I will share what I've done so far, and maybe someone can help me out:

So first of all let us mark $G = \mathbb {Z}_3 \times \mathbb {Z}_3 \times \mathbb {Z}_3$.

Secondly from combinatorial reasons we have $|G| = 3^3 = 27$.

Let us examine the action of $G \mapsto G$ on itself by multiplication from the left.

Let us examine the induced homomorphism $\phi:G \to S_{27}$.

Let $g \in G$. If $g \in$ Ker$\phi$ then we have $\phi(g) = Id_{27}$ and $\phi(g)$ is in the conjugacy class of ().

Otherwise, assume $g \notin$ Ker$\phi$. Let us mark $\phi(g) = \psi_g$ such that:

$\psi_g:G \to G$ upholds for some $\;g^{'} \in G\;$, $\psi(g^{'}) = gg^{'}$.

If we mark $g = \langle a,b,c \rangle$ for some $a,b,c \in \mathbb{Z}_3$ then notice:

$|\langle a,b,c \rangle| \in \{1,2,3\}$ since $(0+0+0),(1+1+1),(2+2+2) \equiv 0 \;(mod 3)$.

So I'm having a hard time continuing from here unfortunately and would love some help...

Answer 1

First, learn about the conjugacy classes in symmetric groups. Any conjugacy class is the set of all permutations with a given cycle type. There are many resources explaining this.

Then, you must show $\phi(g)$ is a product of $3$-cycles for any nontrivial $g$.

Note $g^3=e$ but $g\ne e$. This automatically implies $\phi(g)^3=e$. Thus, every cycle of $\phi(g)$ has length $1$ or $3$, and we can rule out length $1$. (Why?)

The fact that $G=\mathbb{Z}_3^3$ is actually not relevant. In the regular representation of any group, $\phi(g)$ is a product of equal-length cycles. So you don't need to use anything about $\mathbb{Z}_3^3$ other than the fact that it is a finite group.

Answer 2

Consider three $3$-cycles $\gamma_1,\gamma_2,\gamma_3$ in $S_{27}$ with disjoint supports, so they commute. Then the map $f: (k_1+3\mathbb{Z},k_2+3\mathbb{Z},k_3+3\mathbb{Z}) \mapsto \gamma_1^{k_1}\gamma_2^{k_2}\gamma_3^{k_3}$ from $(\mathbb{Z}/3)^3$ to $S_{27}$ is well-defined and is an injective group homomorphism. Its image contains elements of four different conjugation classes, namely products of $r$ disjoint $3$-cycles, for $0 \le r \le 3$.