# Number of homomorphisms between two arbitrary groups

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 ?

• $A_5$ is simple. Dec 20 '18 at 18:26
• So $A_5$ is simple and since order of $A_5$ is greater than order of $S_4$ , It cannot be one-one mapping. So trivial Dec 20 '18 at 18:40
• The question has been answered here already. Dec 20 '18 at 19:15
• List all the functions between the two sets of group elements and count which ones are group homomorphisms. Dec 20 '18 at 19:26
• The exact number of homomorphisms between two arbitrary groups is rarely of much interest, although for some reason it seems to be a popular homework problem. It is of course interesting to know hwether two groups are isomorphic, or whether one is isomorphic to a subgroup or quotient of the other, but the exact number of homomorphisms has no particular significance. Dec 21 '18 at 8:31

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

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
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 $${()}_{S_4}$$.