MHB How many homomorphisms are there from $\mathbb{Z}_4$ to $S_4$?

[mathmari]

Hey! :i

How many homomorphism $f:\mathbb{Z}_4\rightarrow S_4$ are there?

Do we have to find how many permutations of $S_4$ have order that divides $4$ ?

We have 1 identity (order 1), 6 transpositions (order 2), 3 products of two disjoint transpositions (order 2), 6 4-cycles (order 4).

So in total we have 1 + 6 + 3 + 6 = 16 elements of $S_4$ that have order that divides 4, right?

Does this mean that there are $16$ homomorphisms? (Wondering)

 

[I like Serena]

Hey! :i

How many homomorphism $f:\mathbb{Z}_4\rightarrow S_4$ are there?

Do we have to find how many permutations of $S_4$ have order that divides $4$ ?

We have 1 identity (order 1), 6 transpositions (order 2), 3 products of two disjoint transpositions (order 2), 6 4-cycles (order 4).

So in total we have 1 + 6 + 3 + 6 = 16 elements of $S_4$ that have order that divides 4, right?

Does this mean that there are $16$ homomorphisms? (Wondering)

Yep. (Nod)

For a homomorphism we need that $f(a)f(b)=ab$ for any $a,b$.
The structure of $\mathbb Z_4$ is fully identified by its generator $1$.
That means that it suffices indeed to look at $f(1)$, which fully identifies the homomorphism, and in particular verify that $f(1)^4=\text{id}$. (Nerd)

 

[mathmari]

Yep. (Nod)

For a homomorphism we need that $f(a)f(b)=ab$ for any $a,b$.
The structure of $\mathbb Z_4$ is fully identified by its generator $1$.
That means that it suffices indeed to look at $f(1)$, which fully identifies the homomorphism, and in particular verify that $f(1)^4=\text{id}$. (Nerd)

Ok! Thank you! (Yes)