Number of Homomorphisms from $\mathbb{Z}_4$ to $S_4$: A Brief Exploration

[evinda]

Hello! (Wave)

We are given the groups $G_1=\mathbb{Z}_4$ and $G_2=S_4$. We consider the homomorphisms $f: G_1 \to G_2$. Let $k$ be the number from all of these $f$. What is $k \bmod{6}$ equal to ?

How can we find the number of homomorphisms $f$? Could you give me a hint? (Thinking)

 

[I like Serena]

Hello! (Wave)

We are given the groups $G_1=\mathbb{Z}_4$ and $G_2=S_4$. We consider the homomorphisms $f: G_1 \to G_2$. Let $k$ be the number from all of these $f$. What is $k \bmod{6}$ equal to ?

How can we find the number of homomorphisms $f$? Could you give me a hint?

Hey evinda!

We can identify and recognize homomorphisms based on the images of the generators.
$G_1$ is generated by $1$. That is, $G_1=\langle 1 \rangle$.
Once we pick an image for $1$, the images of all other elements are fixed. (Nerd)

If we pick for instance $f(1)=(12)$, then it follows that $f(2)=(), f(3)=(12)$, and we always have $f(0)=()$.
Note that the order of $1$ is $4$.
This order has to be a multiple of the order of the image $(12)$, which has order $2$.
So we've found one homomorphism.

How many possible images do we have for $1$? (Wondering)

And how many of the generated maps that correspond to them are actually homomorphisms? (Wondering)

 

[evinda]

Once we pick an image for $1$, the images of all other elements are fixed. (Nerd)

If we pick for instance $f(1)=(12)$, then it follows that $f(2)=(), f(3)=(12)$, and we always have $f(0)=()$.

How did we find the $f(2), f(3), f(0)$ ?

This order has to be a multiple of the order of the image $(12)$, which has order $2$.

Why does this have to hold? (Thinking)

So we've found one homomorphism.

How do we deduce this? (Worried)

How many possible images do we have for $1$? (Wondering)

How do we find how many possible images for $1$ there are? (Thinking)

And how many of the generated maps that correspond to them are actually homomorphisms? (Wondering)

I don't know so far... (Tmi)

 

[I like Serena]

How did we find the $f(2), f(3), f(0)$ ?

Can we apply the definition of a homomorphism to find them? (Wondering)

What is the definition of a homomorphism? (Wondering)

Why does this have to hold?

Because of what $f(0)$ has to be, based on the application of the definition that we are doing.

How do we deduce this?

By checking if the map we found satisfies the definition of a homomorphism.

How do we find how many possible images for $1$ there are?

How many elements does $S_4$ have? (Wondering)

I don't know so far...

Let's get back to that when we have applied the definition of a homomorphism to the previous questions. (Thinking)

 

[evinda]

Can we apply the definition of a homomorphism to find them? (Wondering)

What is the definition of a homomorphism? (Wondering)

The definition is that $f(u+v)=f(u) \cdot f(v)$, right? (Thinking)

So if $f(1)=(1 2) $, then $f(2)=f(1+1)=f(1)f(1)=(1 2)(1 2)=()$, $f(3)=f(2+1)=f(2)f(1)=()(1 2)=(1 2)$, $f(0)=f(4)=f(3)f(1)=(1 2)(1 2)=()$ right?

Because of what $f(0)$ has to be, based on the application of the definition that we are doing.

I still haven't understood why the order of $1$ has to be a multiple of the order of $(12)$...

By checking if the map we found satisfies the definition of a homomorphism.

So do we have to check the $f$ at each element of its domain seperately to be able to say if it is a homomorphism? (Thinking)

How many elements does $S_4$ have? (Wondering)

$4!$, right?

Let's get back to that when we have applied the definition of a homomorphism to the previous questions. (Thinking)

Ok... (Nod)

 

[I like Serena]

The definition is that $f(u+v)=f(u) \cdot f(v)$, right? (Thinking)

So if $f(1)=(1 2) $, then $f(2)=f(1+1)=f(1)f(1)=(1 2)(1 2)=()$, $f(3)=f(2+1)=f(2)f(1)=()(1 2)=(1 2)$, $f(0)=f(4)=f(3)f(1)=(1 2)(1 2)=()$ right?

Yes.
And note that $f(4)=f(0)=()$. (Nerd)

I still haven't understood why the order of $1$ has to be a multiple of the order of $(12)$...

Suppose we pick $f(1)=(123)$, what will $f(4)$ be? (Wondering)

So do we have to check the $f$ at each element of its domain seperately to be able to say if it is a homomorphism?

According to the definition we have to check every pair of elements in $G_1$ and check if they satisfy the condition.
We can make it a bit easier though.
Suppose $g$ is the single generator of $G_1$, and suppose $g$ has order $m$.
Than we can freely pick $f(g)$ with only the condition that $f(g^m)=f(g)^m$.
What does that mean for the order of $f(g)$? (Wondering)

$4!$, right?

Yep.
So we have 24 candidates for homomorphisms. (Thinking)

 

[evinda]

Suppose we pick $f(1)=(123)$, what will $f(4)$ be? (Wondering)

We will have $f(2)=(3 2 1)$, $f(3)=()$ and $f(4)=(1 2 3)$, right? (Thinking)

Suppose $g$ is the single generator of $G_1$, and suppose $g$ has order $m$.
Than we can freely pick $f(g)$ with only the condition that $f(g^m)=f(g)^m$.
What does that mean for the order of $f(g)$? (Wondering)

Why does it have to hold that $f(g^m)=f(g)^m$ ?

Don't we get from the definition that $f(g)^m=f(g) \cdots f(g)=f(mg)$ ? Or am I wrong? (Thinking)

Yep.
So we have 24 candidates for homomorphisms. (Thinking)

(Nod)

 

[I like Serena]

We will have $f(2)=(3 2 1)$, $f(3)=()$ and $f(4)=(1 2 3)$, right? (Thinking)

Yes.
And since 4=0 we must have f(4)=f(0).
Is that the case?

Why does it have to hold that $f(g^m)=f(g)^m$ ?

Don't we get from the definition that $f(g)^m=f(g) \cdots f(g)=f(mg)$ ? Or am I wrong?

Yes, since we are adding in G1, we must have $f(0)=f(mg)=f(g)^m=()$. (Thinking)

 

[evinda]

Yes.
And since 4=0 we must have f(4)=f(0).
Is that the case?

You said that we always have $f(0)=()$. Why?
Since this holds, it is not the case that $f(4)=f(0)$...

Yes, since we are adding in G1, we must have $f(0)=f(mg)=f(g)^m=()$. (Thinking)

A ok. And from this we get that the order of $f(g)$ divides $m$.
If the order would be a smaller integer than $m$, the order of $g$ would also be smaller, which is a contradiction.
Thus the order of $f(g)$ is $m$, right?

 

[I like Serena]

You said that we always have $f(0)=()$. Why?

It's a property of a homomorphism.
We can prove it by observing that for any $a$ we have $f(a)=f(a+0)=f(a)f(0)$. Therefore $f(0)$ must be the identity. (Nerd)

Since this holds, it is not the case that $f(4)=f(0)$...

Indeed, so with $f(1)=(123)$ it's not an homomorphism. (Thinking)

A ok. And from this we get that the order of $f(g)$ divides $m$.

Correct. (Nod)

If the order would be a smaller integer than $m$, the order of $g$ would also be smaller, which is a contradiction.
Thus the order of $f(g)$ is $m$, right?

Huh?
Didn't we define $m$ as the order of $g$?

 

[evinda]

It's a property of a homomorphism.
We can prove it by observing that for any $a$ we have $f(a)=f(a+0)=f(a)f(0)$. Therefore $f(0)$ must be the identity. (Nerd)

Ok, I see... (Nod)

Indeed, so with $f(1)=(123)$ it's not an homomorphism. (Thinking)

(Nod)

Huh?
Didn't we define $m$ as the order of $g$?

Yes, I wanted to prove that if the order of $f(g)$ would be smaller than $m$, then this would also hold for $g$, which is a contradiction.

 

[I like Serena]

Yes, I wanted to prove that if the order of $f(g)$ would be smaller than $m$, then this would also hold for $g$, which is a contradiction.

Didn't we already see with f(1)=(12) that we have m=4 and order(f(1))=2? (Wondering)

Anyway, which cycle types does $S_4$ have and what are their respective orders? (Wondering)