Find all homomophic images of $S_3$
I'm studying for a test and came across this question. I know that I will have a queestion on the test asking me to determing the homomorphic image of something. A homomorphism is a mapping from one group to another that preserves the operation.
The answer to the above question is $S_3, \mathbb Z_2,$ and ${e}$ but I don't understand how to get there. What is asking when it wants all homorphic images?
On
The question is asking:
For which groups $G$ is there a surjective homomorphism $\phi: S_3 \to G$ ?
By the first isomorphism theorem, the order of $G$ divides the order of $S_3$, which is $6$, and so can only be $1,2,3,6$. The possible groups $G$ are then $C_1$, $C_2$, $C_3$, $C_6$, $S_3$.
$C_1$: take $\phi$ the trivial homomorphism.
$C_2$: take $\phi$ the sign function.
$C_3$: cannot happen because every transposition would be taken by $\phi$ to $1 \in C_3$ since $C_3$ has no elements of order $2$. Thus, $\phi$ would be the trivial homomorphism and so not surjective.
$C_6$: cannot happen because every surjective homomorphism $S_3 \to C_6$ would be injective and so an isomorphism, but $S_3$ is not isomorphic to $C_6$, since $S_3$ is not abelian.
$S_3$: take $\phi$ the identity function.
From the first isomorphism theorem any homomorphic image of a group is isomorphic to a quotient of the group by a normal subgroup.
So in the case of $S_3$ we need only find the normal subgroups of $S_3$. You can check that the only normal subgroups are $A_3$, the trivial subgroup, and $S_3$ itself. So any homomorphic image of $S_3$ must be isomorphic to one of $S_3/\{e\} \simeq S_3$, $S_3/A_3\simeq \mathbb{Z}_2$, or $S_3/S_3 \simeq \{e\}$.
The general strategy for determining all homomorphic images of a group $G$ is to find all normal subgroups of $G$ and determine the groups obtained by taking the quotient of $G$ by each normal subgroup. By the first isomorphism theorem this gives a complete list of the homomorphic images of $G$ up to isomorphism.