Number of homomorphisms between two arbitrary groups

Question

How many homomorphisms are there from $A_5$ to $S_4$ ?

This is how I tried to solve it.

If there is a homomorphism from $A_5$ to $S_4$ , then order of element of $S_4$ should divide the order of its preimage. Now what are the possible order of elements in $S_4$.1,2,3 and 4. Since $A_5$ contains (12345), which is of order 5.. what could be image of (12345). Definitely Identity element which is of order 1. Similarly all 5 cycles must be mapped to identity. There are 24 elements of 5 cycles. 24 elements out of 60 are mapped to identity .. now only two types of homomorphisms are possible. either 30:1mapping or 60:1 mapping. Consider (12)(34) which belongs to $A_5$. It's image can be element of order 2 or identity.there are 15 elements of order 2 . suppose these 15 elements are mapped to some element 'g' of order 2 of $S_4$, you need another 15 elements to get mapped to 'g' to have 30 :1 mapping. Other type of elements left in $A_5$ is of order 3. None of them can be mapped to g. hence 15 elements of order 2 should be mapped to identity .. so , (24+15=39) elements mapped to identity.As mentioned earlier it should be 30 :1 or 60:1 mapping. So it must be 60:1 mapping.Hence a trivial homomorphism. Answer is 1.

I wanted to know is there any other technique which can be used to find number of homomorphism in the above question ? In general, how to find number of homomorphism between any two arbitrary groups ?

Answer 1

Suppose $f:A_5 \to S_4$ be a homomorphism. Then $\ker f$ is a normal subgroup of $A_5$. But $A_5$ is simple, so $$\ker f \in \Big\{ \{e\},A_5\Big\}$$

• $\ker f=\{e\}$ implies $$A_5/\{e\} \sim f(A_5)$$ and so $f(A_5)$ is a subgroup of order $60$ in $S_4$, which is not possible in $S_4$.
• $\ker f=A_5$ implies $f$ is trivial

Hence $$\Big\vert\{f \;\vert \;f:A_5 \to S_4 \;\text{is a homomorphism} \}\Big\vert=1$$

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For finding homomorphism $f$ for arbitraay two groups, use the following facts:

• $\vert f(g) \vert$ divides $\vert g \vert$ where $g$ belong to the domain with $\vert g \vert < \infty$ [this is useful for finite groups]
• $f(g^n)=[f(g)]^n$
• List all normal subgroups of domain and use first isomorphism theorem

Answer 2

You can rely on another property of $A_5$, other than its simplicity, to get that the only homomorphism from $A_5$ to $S_4$ is the trivial one, namely the fact that $A_5$ has no subgroups of order $30$, $20$ and $15$ (see e.g. here). In fact, a homomorphism $\varphi\colon A_5\to S_4$ is equivalent to a $A_5$-action on the set $X:=\{1,2,3,4\}$. By the Orbit-Stabilizer theorem and the fact that the set of orbits forms a partition of $X$, the stabilizers can only have orders $60/k$, for $1\le k\le 4$; but the stabilizers are subgroups of the group which acts, and hence, by the abovementioned property of $A_5$, the only option $k=1$ is actually allowed (for every $i=1,2,3,4$). So, all the stabilizers must coincide with the whole $A_5$ and the only sought homomorphism has kernel $\bigcap_{i=1}^4{\rm{Stab}}(i)=A_5$, which precisely means that all the elements of $A_5$ are mapped to ${()}_{S_4}$.