-1

My question is :

Let us assume that we have to find the number of homomorphisms from $S_n \to D_{2n}$ when $n > 3$

How to prove that the elements of the form $xyx^{-1}y^{-1}$ always belong to the kernel.

So all the even number of $2$-cycles can be written in the form $xyx^{-1}y^{-1}$ and they belong to the kernel. [which is what I know as $A_n$ is the only normal subgroup of $ S_n$ except when $n \ne 4$,but I can't use this result as I will have to prove this using sylows theorem which we haven't been taught so can't be used in exam ]

Derek Holt
  • 90,008
Guria Sona
  • 1,535
  • 1
    (i) Just to be clear, is $D_{2n}$ the dihedral group of order $2n$? [Some people use the symbol for the one of order $4n$.] (ii) What you say is false for $n=3$. (iii) In the final paragraph do you mean "Any product of an even number of $2$-cycles is (? a product of elements) of this form? (iv) There are proofs that $A_n$ is the only proper non-trivial normal subgroup of $S_n$ which don't use Sylow. – ancient mathematician Nov 03 '20 at 07:48
  • 1)$D_2n$ is the group of order $2n$. 2) when $n=3$ it is isomorphic (3) any two cycle of the form $(ab)(cd)=xyx^{-1}y^{-1}$ – Guria Sona Nov 03 '20 at 07:50
  • 2
    Yes, so the result is wrong when $n=3$. – Derek Holt Nov 03 '20 at 08:14
  • Do you have any help as to how I am going to see the result when $n \ne 3$. – Guria Sona Nov 03 '20 at 08:18
  • 4
    Well when $n$ is odd the permutation $(ab)(cd)$ maps to the product of two commuting elements of order $1$ or $2$, and also it maps to a product of conjugate elements, so either both map to the identity or both map to the same involution. Hence the homomorphism is either trivial or takes every transposition to the same involution $\tau$ and we now know everything. – ancient mathematician Nov 03 '20 at 08:46
  • And essentially the same argument will deal with $n$ even (still $n> 4$) once you establish the conjugacy classes of involutions in $D_{2n}$. – ancient mathematician Nov 03 '20 at 10:49
  • @ancientmathematician thank you for the answer that was what I was precisely looking for. – Guria Sona Nov 03 '20 at 17:01
  • I have one doubt in my mind if f(ab)f(cd)f(ab)=f(cd).How can we conclude that f(ab)=f(cd)? – Guria Sona Nov 03 '20 at 17:02
  • Look at the commuting involutions in $D_{2n}$. – ancient mathematician Nov 03 '20 at 19:26
  • 1
    I was just wondering if $\varphi(ab)-> s $ and $\varphi(cd)-> sr^{n/2}$.Then $s.sr^{n/2}=sr^{n/2}s$, and both of them are of order $2$.This can hold true when $n$ is even though.. – Antimony Nov 04 '20 at 03:00

1 Answers1

4

I really like @ancientmathematician's argument above but if you want here's another argument.

Suppose $N$ is the kernel of your homomorphism, then $N$ has index dividing $2n$, so all $2n$th powers are in $N$, in particular, those elements whose order is coprime to $2n$ are in $N$. Now choose odd number $k\leq n$ coprime to $2n$ (the only time you can't do this is when $n=3$ which as the others mentioned is a counterexample, otherwise choose $k=n-1$ or $n-2$ works) then all $k$-cycles by the above are in $N$.

All $k$-cycles for $k$ odd generate $A_n$ so $A_n\subset N$. If you don't yet know this see here or here And it's clear that all commutators are in $A_n$.

August Liu
  • 464