Determine if H is a normal subgroup of G - faster way than finding cosets?

Question

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

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.

Answer 1

If $xhx^{-1} \in H$ for all $h\in H$ and for all $g\in G$ the subgroup $H$ is normal.

Answer 2

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.

Answer 3

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!

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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.