Question

Find all homomorphic images of $S_3$.

Short answer

The homomorphic images of $S_3$are$\left\{0\right\},{\mathbb{Z}}_2,$ and $S_3$.

Step-by-step solution

Referring to the First IsomorphismTheorem

Theorem 8.20

Let $f:G\to H$be a surjective homomorphism of a group with kernel K. Then the quotient group is$G/K$ isomorphic to H.

Finding the homomorphic images of S3

Consider a map $f:S_3\to G$.

Then, byFirst Isomorphism Theorem,$S_3/K\cong G$.

Now, the order of $S_3/K$= $\left|S_3\right|/\left|K\right|$.

As we know$\left|S_3\right|$=6, the order of the image must be a divisor of 6, i.e., it can be 1, 2, 3 and 6.

Examining all the four possibilities for the given order as:

$f:S_3\to G$

1) Suppose the order is 1.

Then, it represents a trivial group.

Therefore, $f:S_3\to \left\{0\right\}$.

By $f\left(a\right)=0,$for $a\in S_3$.

So,$\left\{0\right\}$is a homomorphic image of $S_3$.

2) Suppose the order is 6.

Then, consider $f:S_3\to S_3$.

Here, the kernel is trivial because both the groups have the same order.

So, $S_3$is a homomorphic image of $S_3$.

3) Suppose order is 3.

As we already know that, the only normal subgroup of$S_3$with order 3 is$A_3$.

As we know from First Isomorphism Theorem that${S_3/A}_3$should be isomorphic to the homomorphic image; so its order should be 2.

Since there isonly one group of order two, which is${\mathbb{Z}}_2$.

Therefore,${\mathbb{Z}}_2$is a homomorphic image of $S_3$.

4) Suppose order is 3.

Then, $S_3$does not have any normal group of order 2.

Therefore, it does not have any homomorphic image of this order.

Therefore, the homomorphic images of$S_3$ are$\left\{2\right\},{\mathbb{Z}}_2,$and $S_3$.