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I am willing to find the total number of group homomorphism from $U(8):=\{1\leq r\leq 8: (r, 8)=1\}$ to $S_8$. Now we know that $U(8)\cong \mathbb Z_2\times \mathbb Z_2$. Hence the number is same as $\#Hom(\mathbb Z_2\times \mathbb Z_2, S_8)$.

Here $\mathbb Z_2\times \mathbb Z_2=\langle (1, 0), (0,1): (1,0)^2=(0,1)^2=(1,1)^2=(0,0)\rangle $. SO I think only thing we need to bother the possibilities of $f(1,0), f(0,1)$ such that order of $f(1,0)f(0,1)$ be 2 only where $f\in Hom(\mathbb Z_2\times \mathbb Z_2, S_8)$.

Here $|f(1,0)|, |f(0,1)|$ have possibilites 1, 2.

If $|f(1,0)|=1$ then for $f(0,1)$ we get total possibilities $\{1+\binom{8}{2}+\frac{1}{2!}\binom{8}{2}\binom{6}{2}+\frac{1}{3!}\binom{8}{2}\binom{6}{2}\binom{4}{2}+\frac{1}{4!}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} \}=764$.

But if $f(1,0)$ has order 2 then ?

Got stuck and confused. Can someone enlighten me please ??

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  • $\begingroup$ Opps !! So sorry for the typo. Editing it :-P $\endgroup$ – Anjan3 Sep 13 '15 at 17:14
  • $\begingroup$ Someone Please help me! :-( $\endgroup$ – Anjan3 Sep 13 '15 at 17:29
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Thanks to @MiloBrandt for pointing out a huge flaw with the first version of this.

How many elements of order $\leq2$ are there in $S_8$ to send $(1,0)$ to? For each, how many elements of order $\leq2$ that commute with the image of $(1,0)$ to send $(0,1)$ to?

  • $\binom{8}{0}$ identity to send $(1,0)$ to, with $\binom{0}{0}\left(\binom{8}{0}+\binom{8}{2}+\frac{1}{2!}\binom{8}{2}\binom{6}{2}+\frac{1}{3!}\binom{8}{2}\binom{6}{2}\binom{4}{2}+\frac{1}{4!}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}\right)$ options for $(0,1)$
  • $\binom{8}{2}$ 2-cycles to send $(1,0)$ to, with $\left(\binom{1}{0}+\binom{1}{1}\right)\left(\binom{6}{0}+\binom{6}{2}+\frac{1}{2!}\binom{6}{2}\binom{4}{2}+\frac{1}{3!}\binom{6}{2}\binom{4}{2}\binom{2}{2}\right)$ options for $(0,1)$
  • $\frac{1}{2!}\binom{8}{2}\binom{6}{2}$ products of two disjoint 2-cycles to send $(1,0)$ to, with $\left(\binom{2}{0}+\binom{2}{1}+\binom{2}{2}\right)\left(\binom{4}{0}+\binom{4}{2}+\frac{1}{2!}\binom{4}{2}\binom{2}{2}\right)$ options for $(0,1)$
  • $\frac{1}{3!}\binom{8}{2}\binom{6}{2}\binom{4}{2}$ products of three disjoint 2-cycles to send $(1,0)$ to, with $\left(\binom{3}{0}+\binom{3}{1}+\binom{3}{2}+\binom{3}{3}\right)\left(\binom{2}{0}+\binom{2}{2}\right)$ options for $(0,1)$
  • $\frac{1}{4!}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$ products of four disjoint 2-cycles to send $(1,0)$ to, with $\left(\binom{4}{0}+\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4}\right)\binom{0}{0}$ options for $(0,1)$

In the counting above, I leave it to the reader to understand the counting of options to send $(1,0)$ to. When counting where to send $(0,1)$ to, we first count how many of the 2-cycles in the image of $(1,0)$ that will be in the image of $(0,1)$, and then count how many elements of order $\leq2$ there are that are disjoint from the image of $(1,0)$. Multiply, and add.


This is still not enough. It has left out sending the two generators to cycle combinations that overlap while still commuting. For instance, $(12)(34)$ commutes with $(13)(24)$. Perhaps there's not much more to count and add in, but I'm going to leave it for now.


This accounting of subgroups of $S_8$ tells us the ultimate answer to your question is (counting from the bottom of the table) $$\begin{align}&1+(105+28+210+420)\cdot3\\&\phantom{1}+(315+630+210+420+70+1260+210+630+420+1260+1260)\cdot\binom{3}{2}\cdot2!\end{align}$$ which is $42400$. My accounting above only gets up to $21820$.

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  • $\begingroup$ This doesn't work; the two elements you send $(1,0)$ and $(0,1)$ to need to commute. So you can't take $(1,0)$ to $(1 2)$ and $(0,1)$ to $(23)$ since then $f(1,1)=f(1,0)f(0,1)=(1 2 3)\neq f(0,1)f(1,0) = (3 2 1)$. $\endgroup$ – Milo Brandt Sep 13 '15 at 17:11
  • $\begingroup$ I did the same sir but some problems arrived. For example if $f(1,0)=(45)$ and $f(0,1)=(5,6)$, then $f(1,1)=f(1,0)f(0,1)=(45)(56)=(456)$ which is not of order 2 whereas $|f(1,1)|=2$. In this way we can tell that $f(1,0), f(0,1)$ should not share any common elements. Else contradiction will arrive. And hence the desired number will become less than $N^2$ you mentioned above. Here is my trouble. How shall I delete the redundant cases ? :-( $\endgroup$ – Anjan3 Sep 13 '15 at 17:13
  • $\begingroup$ @MiloBrandt Thanks: hopefully addressed now. $\endgroup$ – alex.jordan Sep 13 '15 at 17:36
  • $\begingroup$ @MiloBrandt Never mind. See my note at the end. $\endgroup$ – alex.jordan Sep 13 '15 at 17:44
  • $\begingroup$ @alex.jordan Agh! I was trying to find a more elegant solution, but that observation makes my approach fail too. Hmm.... this isn't as easy as I'd hoped. $\endgroup$ – Milo Brandt Sep 13 '15 at 17:45

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