3
$\begingroup$

I am trying to show that $D_8$ is not a normal subgroup of $S_4$, I have shown that $D_8$ is Subgroup of $S_4$.Then, let $r=\langle(1234)\rangle$ and $s =\langle(14)(23)\rangle$.Then, we see that $D_8$ is a subset of $S_4$ and since it a group so $D_8$ is a subgroup of $S_4$.

Proving that $D_8$ is not a normal subgroup. In $D_8$ there are two elements of order $4$.So, we take $\sigma (1234) \sigma^{-1}$ which should be in $H$.Now conjugacy preserves the cycle structure so we choose $\sigma$ in such a way so that conjugate element does not belong to $H$.Choosing such a cycle is possible as there are $6$ 4 cycles in $S_4$ and there are only two of them in $D_8$.Since I have shown it for one element, that the conjugate is not in $D_8$.Do I need to show it for the other elements?

$\endgroup$

2 Answers 2

3
$\begingroup$

A subgroup $H$ of a group $G$ is normal if and only if for all $g\in G$ and $h\in H$, $ghg^{-1}\in H$.

Since you showed that there exists $g\in S_4$ such that $g(1234)g^{-1}\notin D_8$, it is sufficient to conclude that $D_8$ is not a normal subgroup of $S_4$.

$\endgroup$
0
$\begingroup$

If we view the elements $1,2,3,4$ as vertices of a square, and consider the dihedral group of that square, this gives us one copy of $D_8$. We can obtain other copies by reordering.

There are $6$ ways to order $4$ vertices (assuming we view cyclic shifts such as $1,2,3,4$ and $2,3,4,1$ as the same ordering), resulting in $3$ distinct copies of $D_8$ (because "opposite" pairs of orderings such as $1,2,3,4$ and $1,4,3,2$ result in the same group; just flip the square upside-down to go from one ordering to the other).

These three distinct copies of $D_8$ are conjugate to each other, e.g. because they are Sylow $2$-subgroups (they have the correct order), and all Sylow subgroups of a given order are conjugate. In particular they are not normal.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .