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First:

Let $G=S_4$ - the group of all permutations of {1,2,3,4}.

I must prove $N=\{1,(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\}$ is a subgroup of $G$ - this is easy, each transposition is its own inverse, so the inverse of $(1,2)(3,4)$ is it backwardss, that is $[(3,4)(1,2)](1,2)(3,4)$ you can use the fact that disjoint cycles commute (this is a bit wooly....)

Anyway that's easy enough but anyway it is a subgroup. (The Klein 4 group if I have recognised it correctly)

Second:

"It is actually a normal subgroup" (I don't quite see what's good about these other than that $|G/N|=|G|/|N|$ because of it)

We have $|G/N|=|G|/|N|=24/4=6$ By a proposition in the book we have either $G/N\simeq C_6$ - the cyclic group of order 6, or $G/N\simeq D_6$ which is the "Dihedral group" on 6 points (and is of order 12), it consists of all permutations that can be obtained by reflecting and rotating a hexagon.

but there's no element of order 6 in $S_4$ (can't prove this, not sure how, but surely it is obvious!)

However the regular hexagon has an element of order 6, a rotation of $\frac{2\pi}{6}$ .... so what on Earth is going on...

I believe there should be a really elegant way to say this definitively in one swing, with no effort.

I am questioning my assumption that $G/N\subset G$ ... what does that define? I assumed "left coset"

Source: Rings Fields and Groups, R B J T Alenby - page 229

duplicate of Showing an Isomorphism between question group of $S_4$ and $D_6$ but I am much further away from the answer than him. I've just realised what I thought made no sense while writing this, I hope I still receive useful information despite this...

(that question is a suggested duplicate of something to do with a kernel, I am not sure what a kernel is in this context, so please answers without using kernels)

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Alec Teal
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  • I have inserted \simeq to denote isomorphy. – AlexR Feb 02 '14 at 23:03
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    If the book offers $G/N \simeq D_6$ as a possibility, it probably designates the dihedral group of a triangle, with six elements, by that. (That group is also commonly known as $S_3$, the symmetric group on three symbols.) – Daniel Fischer Feb 02 '14 at 23:17
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    $D_6$ as in your notation has $12$ elements, $G/N$ has only $6$. Some authors use $D_{2n}$ to denote the dihedral group of order $2n$ that acts on $n$ vertices, while others use $D_n$. – Pedro Feb 02 '14 at 23:18
  • @PedroTamaroff so it is not $C_6$ because that means there's a permutation of 4 things with order 6? Is that it? – Alec Teal Feb 02 '14 at 23:39
  • "I don't quite see what's good about these other than that..." It is always true the coset space $G/N$ has cardinality $|G|/|N|$ (in the finite case). But it can be given a "natural" group structure, namely $aNbN:=(ab)N$ iff $N$ is normal. – Pedro Feb 02 '14 at 23:53
  • @PedroTamaroff can you give an example where this isn't the case? (I mean where H is not normal but $g^n\in H$ for any $g\in G$ - or where $|G/N|\not=|G:N|=n$) – Alec Teal Feb 03 '14 at 01:04
  • $|G/N|$ always equals $|G|/|N|$ in the finite case, since $G/N$ denotes the collection of all cosets of $N$ in $G$, and there are as many left cosets as there are right cosets. – Pedro Feb 03 '14 at 01:13

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