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How many group homomorphisms are there from the dihedral group $D_4$ with eight elements to $S_{3}$?

I am really confuse about this type of problem. How do we find all the group homomorphisms between them, I don't even know how to define the function between them. Can anyone tell me about that?

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Let $f:D_4\to S_3$ be a group homomorphism .By First Isomorphism Theorem;

$D_4/\ker f \cong \text{Image }f$.

Possible orders of $\text{Image }f=1,2,3,6$ .Now $|\text{Image }f|\neq 3,6$

If $|\text{Image }f|=1\implies f(a)=e \forall a\in D_4$ where $e$ denotes the identity mapping in $S_3$.

If $|\text{Image }f|=2\implies |\ker f|=4$ .

Take $\text{Image }f=\{e,(12)\}/\{e,(23)\}/\{(e,13)\}$. Also $D_4$ has three normal subgroups of order $4$ denote them by $T_1,T_2,T_3$.

So we can have $3\times 3$ choices for homomorphisms when $|\text{Image }f|=2$

So we have $9+1=10$ group homomorphisms.

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  • $\begingroup$ Is $|\text{Image }f|\neq 3,6$ because of the order of $D_{4}$ is 8? $\endgroup$ – 梁楷葳 Nov 6 '16 at 16:23
  • $\begingroup$ I don't understand this: $|\text{Image }f|=2\implies |\ker f|=4$. Can you explain that to me please? $\endgroup$ – 梁楷葳 Nov 6 '16 at 16:41
  • $\begingroup$ Yes sure; if $\text{|Image}|=3$ then $|\ker f|=\frac{8}{3}$ ;Is it possible;Also $\frac{|D_4|}{|\ker f|}=\text{|Image}|$ @梁楷葳; $\endgroup$ – Learnmore Nov 6 '16 at 17:34
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Hint: The dihedral group with eight elements is the set of symmetries of a polygon with 4 vertices, symbolized $D_4$. The group is generated by a rotation $r$ of order 4, and a reflection, $s$ of order 2. That is, $D_4=\langle r, s\rangle$. So any group homomorphism from $D_4$ into $S_3$ must assign $r$ to an element whose order is divisible by 4, and $s$ to an element whose order is divisible by 2. These selections will determine the homomorphism completely.

So now the question becomes - how many elements of order 1,2, or 4 are there in $S_3$?

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  • $\begingroup$ order 1: 1 element order 2: 3 elements order 3: 2 elements No elements with order 4 in $S_{3}$ right? $\endgroup$ – 梁楷葳 Nov 6 '16 at 16:54

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