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$G = S_4$, $H = \{(1),(12)(34),(13)(24),(14)(23)\}$.

I just did it the long way, and found H to be normal. Is there a better way than finding left and right cosets? I don't want to spend this kind of time during a test if it comes up again.

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    $\begingroup$ You can show that $H$ is normal if you can show it is the kernel of a homomorphism. $\endgroup$ – John Douma Nov 25 '15 at 4:37
  • $\begingroup$ i definitely recognize all these words from parts of the textbook i was supposed to read but haven't gotten around to learning properly yet. this is probably the way i was meant to do, it if the question wasn't meant to be one of the tedious exercises that so often appear at the start of assignments. $\endgroup$ – Ryan McCabe Nov 25 '15 at 5:03
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    $\begingroup$ Because it's now missing but it looked like a great answer (and I use previously answered questions to get through my homework all the time): Someone said that Sylow theorems solve this quickly and simply. I don't remember the precise details but check that route, too, future lost souls. $\endgroup$ – Ryan McCabe Nov 25 '15 at 7:04
  • $\begingroup$ @RyanMcCabe it was my answer that was deleted, because later I found out that it was not a sylow $2$ group , I assumed it wrongly though I got the answer correct. Actually your group is Klein $4$ group which is a normal subgroup of $S_4$, but i didn't post that as an answer. Sorry for the wrong answer. $\endgroup$ – Kushal Bhuyan Nov 26 '15 at 6:55
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In the case of $G = S_n$, two elements $g_1, g_2$ are conjugate (i.e. $g_1 = g g_2 g^{-1}$ for some $g \in G$) if and only if they have the same cycle type; that is, they are composed of the same number of cycles of the same lengths. In this case, $(1)$ has cycle type $(1,1,1,1)$, the other three elements of $H$ have cycle type $(2,2)$, and there are no other elements of $G$ with either of those cycle types. So every conjugate of an element of $H$ must be in $H$ itself, which is equivalent to saying that $H$ is normal.

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If you know a set of generators for G and a set of generators for H, then, as long as G is finite, you can check that H is normal by verifying that conjugating each generating element of H by each generating element of G in turn gives you an element which is again in H. If G or H are infinite, then you need to make sure that the respective generating sets are closed under inversion (i.e. $x \in S \implies x^{-1} \in S$). If you don't know generating sets for G or H, you can also use all of the elements.

For instance, in your example, there's a generating set of $S_4$ with only two elements (I hope you know this!). Just using that alone brings the amount of computation needed way down!


This follows from a couple of facts:

  1. If a finite group K is generated by $k_1$,...,$k_n$, then any element of K can be written as a product of some $k_i$'s (repeats allowed). You don't need inverses because $x^{-1} = x^{o(x)-1}$, and $o(x)$ is finite.
  2. A subgroup H is normal in G if $g^{-1}Hg = H$ for all $ g \in G$.
  3. $(g_1g_2)^{-1}h(g_1g_2) = g_2^{-1} (g_1^{-1}h g_1) g_2$, so you can expand conjugation by a product to repeated conjugation.
  4. $g^{-1}(h_1 h_2)g = g^{-1}h_1g \ast g^{-1}h_2g$, so you can expand conjugation of a product to a product of conjugates.
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  • $\begingroup$ okay this sounds really useful and of broad application. is that just some kind of corollary of normal groups? i could just take any two 2-cycles from H and check by conjugating them with (12) and (1234) to make a grand total of 4 calculations instead of 80? $\endgroup$ – Ryan McCabe Nov 25 '15 at 20:57
  • $\begingroup$ @RyanMcCabe: I've edited my answer with the facts you need to prove my statement. Let me know if you need some more clues/explanation! $\endgroup$ – yatima2975 Nov 26 '15 at 6:48
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    $\begingroup$ Just to add: even in the infinite case, you can get away with the generating set for $H$ not being closed under inversion, because $g h^{-1} g^{-1} = (g h g^{-1})^{-1}$. Good catch about $G$ though, @yatima2975--I would have assumed it's not necessary, but I see that it is. $\endgroup$ – Ravi Fernando Nov 26 '15 at 19:13
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If $xhx^{-1} \in H$ for all $h\in H$ and for all $g\in G$ the subgroup $H$ is normal.

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  • $\begingroup$ thanks but this one isn't faster, i'm still running the subgroup through every other member of the group. $\endgroup$ – Ryan McCabe Nov 25 '15 at 4:56
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    $\begingroup$ Like Ravi says, conjugation doesn't change the cycle type. Since $H$ has all the elements which are two transpositions i.e. this cycle type, it will always be the case that $xhx^{-1}$ will still be in $H$. $\endgroup$ – CPM Nov 25 '15 at 8:41

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