There is a similar problem that $\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$ to $S_{4}$. But I still confuse,

This is what I tried:

Let $f$ be homomorphism from $\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z \to S_{3}$

then $\Bbb Z_2\times \Bbb Z_2/\ker f \cong \text{Image} f$

$|\text{Image} f |=1,2,4$

If $|\text{Image} f |=1$,

$f(a)=e$ ,for all $a\in \Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$, so that's 4 grooup homomorphisms,

since there are 4 elements in $\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$ (Is that right?)

can anybody solve this problem more detail, I stuck on this problem for two hours

  • $\begingroup$ You got $f(a) = e$ for all $a$. That means there's no choice for $f$ - there's only one group homomorphism in that case. Now you need to consider the cases $|\operatorname{Img}(f)| = 2, 4$. $\endgroup$ – Dustan Levenstein Nov 6 '16 at 20:16
  • $\begingroup$ The image of $f$ will be a subgroup of $S_3$. As such, the size of the image of $f$ must divide the size of $S_3$, which is $6$. Hence, the image of $f$ can't have $4$ elements. $\endgroup$ – Kaj Hansen Nov 6 '16 at 20:18
  • $\begingroup$ @KajHansen So we only need to discuss $|\operatorname{Img}(f)| = 2$? $\endgroup$ – Wade Nov 6 '16 at 20:20
  • $\begingroup$ That's correct. Since $\mathbb{Z}_2 \subset S_3$, consider homomorphisms $\mathbb{Z}_2 \times \mathbb{Z}_2 \rightarrow \mathbb{Z}_2$. $\endgroup$ – Kaj Hansen Nov 6 '16 at 20:31
  • $\begingroup$ So, If $|\operatorname{Img}(f)| = 2$, $f(a)=(13),(12),(23)$, so there are $4 \times 3 $ group homomorphisms? Is that right?@KajHansen $\endgroup$ – Wade Nov 6 '16 at 20:39

Since $4$ does not divide $6$ there are no injections into $S_3$. Let $a,b$ be the generators of $\Bbb Z_2 \times \Bbb Z_2$. Then there are three possibilities for the cyclic kernel of order $2$: $\langle a \rangle, \langle b \rangle,$ and $\langle ab \rangle$. There are three elements of order $2$ in $S_3$, combining these with the possible kernels (and mapping the other two elements to the same image) we obtain $9$ homomorphisms, together with the trivial homomorphism this gives us a total of $10$ homomorphisms.

  • $\begingroup$ There are four possibilities for the image of each generator, but we are constrained to choices that commute. $\endgroup$ – hardmath Nov 7 '16 at 13:07
  • $\begingroup$ @hardmath Why four? $\endgroup$ – Marc Bogaerts Nov 7 '16 at 13:07
  • $\begingroup$ The identity plus the three transpositions. $\endgroup$ – hardmath Nov 7 '16 at 13:08
  • $\begingroup$ The identity and one of the transpositions map to the identity, the other two map to the same element. $\endgroup$ – Marc Bogaerts Nov 7 '16 at 13:10
  • $\begingroup$ To clarify, I also get $10$ homomorphisms. If $a $ goes to the identity, then $b $ can map to any of the four possible images. If $a $ maps to a transposition, then there are two choices for the image of $b $. $4+3\cdot 2 = 10$. $\endgroup$ – hardmath Nov 7 '16 at 13:18

One way to count the group homomorphisms $f:\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z \to S_3$ is to focus on the images of generators $a=(1,0)$ and $b=(0,1)$.

Since $a,b$ are both of order $2$, the images $f(a),f(b)$ must have orders that are divisors of $2$. Therefore each image must be the identity in $S_3$ having order $1$, or (since $S_3$ is permutations of three items) a transposition (having order $2$). Thus four choices in all for each generator image.

But the choices of $f(a),f(b)$, although they fully determine the map $f$, are constrained by the abelian property of $\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$ to satisfy $$f(a)f(b)=f(b)f(a)$$

One way to count these is to subtract from the $4\cdot 4 = 16$ those possibilities for which $f(a)$ and $f(b)$ would not commute. There are six such non-commuting choices, picking one transposition for $f(a)$ and a different transposition for $f(b)$.

Thus we count $16-6=10$ possible homomorphisms, in agreement with the count given by Bogaerts Marc.


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