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Section 1.9 Normal subgroups

Let us go back to the idea of cosets. We know that the cosets are not subgroups of \(G\text{,}\) except for the one that includes the identity element. What we want to do now is define a group structure on the “set of cosets”. That is, the elements of the new group will be the cosets themselves.

To get a group structure, given two cosets, we want to be able to “multiply them”.

Definition 1.9.1. Product of subsets.

Given any two subsets \(S_1\) and \(S_2\) of \(G\text{,}\) we define the product of \(S_1\) and \(S_2\) as being the subset

\begin{equation*} S_1 S_2 = \{ s_1 s_2 ~|~s_1 \in S_1 \text{ and } s_2 \in S_2 \}. \end{equation*}

That is, we multiply each element of \(S_1\) with each element of \(S_2\text{.}\)

Now we want to give the set of cosets a group structure under this operation of multiplication of cosets. It turns out that this is only possible if the left cosets of \(H\) are equal to its right cosets. Or, equivalently, if all elements conjugate to elements of \(H\) are also in \(H\text{.}\) So we give such subgroups a special name:

Definition 1.9.2. Normal subgroups.

A subgroup \(H\) of \(G\) is called normal if \(H g = g H\) for all \(g \in G\text{.}\) Equivalently, it is normal if \(g H g^{-1} = H\) (as sets) for all \(g \in G\text{.}\) Such subgroups are also called self-conjugate, since given any element \(h \in H\text{,}\) all its conjugate elements are also in \(H\text{.}\)

In fact, a subgroup is normal if and only if for all \(h \in H\text{,}\) all the elements conjugate to \(h\) are also in \(H\text{.}\) That is, \(g H g^{-1} \subseteq H\) for all \(g \in G\text{.}\) This may seem like a slighly weaker condition than \(g H g^{-1} = H\) for all \(g \in G\text{,}\) but in fact the latter is implied by the former for any group, as one can prove. Therefore, to prove that a subgroup is normal, it is sufficient to show that all elements conjugate to elements of \(H\) are also in \(H\text{,}\) which is often taken as the definition of a normal subgroup.

Prove that \(g H g^{-1} \subseteq H\) for all \(g \in G\) implies that \(g H g^{-1} = H\) for all \(g \in G\text{.}\)

This is also rather clear. Suppose \(H\) is normal. Pick a \(h \in H\text{.}\) Then, for any \(g \in G\text{,}\) \(g h g^{-1} \in H\text{,}\) and hence the conjugacy class of \(h\) is included in \(H\text{.}\) Since this must true for all \(h \in H\text{,}\) we conclude that \(H\) is the union of the conjugacy classes of elements in \(H\text{.}\)

Conversely, if \(H\) is the union of conjugacy classes, then we know that the elements conjugate to any element in \(H\) are also in \(H\text{,}\) and hence \(H\) is normal.

This gives another way of proving that a subgroup is normal: it will be if and only if it "does not break conjugacy classes", i.e. is the union of conjugacy classes.

Before we study examples of normal subgroups, let us introduce one more piece of jargon:

Definition 1.9.5. Simple groups.

A group that does not have proper normal subgroups is called simple.

Simple, isn't it? All right, let us now look at some examples!

Let us show explicitly that the alternating group \(A_3\) is a normal subgroup of \(S_3\text{.}\) Recall from Definition 1.8.14 that \(A_3\) consists of the even permutations \(\pi_1 = (1)(2)(3), \pi_5 = (132)\) and \(\pi_6 = (1 2 3)\text{.}\) To show that it is normal, one could show that all left and right cosets agree. But let us avoid such lengthy calculations and simply argue that the elements of \(g H g^{-1}\) must be in \(H\) for all \(g \in G\text{.}\) So what we need to show is that all permutations in \(g H g^{-1}\) are even, in which case they are in \(H = A_3\text{.}\) Assume that \(h \in H\) is even. If \(g\) is even, then \(g h g^{-1}\) is certainly even. If \(g\) is odd, then \(g^{-1}\) is also odd, and hence \(g h g^{-1}\) is even. Therefore \(H\) is a normal subgroup. In fact, this argument works for all alternating subgroups \(A_n \subset S_n\text{,}\) which are all normal.

All subgroups of abelian groups are necessarily normal, since \(g h g^{-1} = g g^{-1} h = h\) for all \(g \in G\) and \(h \in H\text{.}\) Equivalently, each element in an Abelian group constitutes a conjugacy class by itself, and hence all subgroups are unions of conjugacy classes. Thus all subgroups are normal. As an example, the group \(H = \{1,-1\}\) under multiplication is a normal subgroup of \(G = \{1,i,-1,-i \}\text{.}\)

The centre \(Z(G)\) of a group \(G\) is always normal. Indeed, for all \(g \in G\) and \(h \in Z(G)\text{,}\) we have \(g h g^{-1} = g g^{-1} h = h\text{,}\) since \(h\) must commute with all elements of \(G\text{.}\) Equivalently, each element in the centre of a group constitutes a conjugacy class by itself, and hence the centre of a group is always a union of conjugacy classes. Therefore it is normal.

Given a direct product \(G_1 \times G_2\text{,}\) the subgroups \((e, G_2)\) and \((G_1,e)\) are normal. Indeed, it is easy to check that these subgroups are self-conjugate.

Another interesting example of a normal subgroup is the subgroup \(C_0\) of the \(3 \times 3\) Rubik's cube group consisting of all moves that leave the position of every block fixed but can change the orientation of the blocks. Why is it normal? Well, if you first do an arbitrary move, than do a move in \(C_0\) which only changes the orientiation of the blocks, and then undo the first move, you will end up with all the block in their original position, but with their orientation potentially changed. In other words, you will end up with an element of \(C_0\text{.}\) Therefore we conclude that \(C_0\) is a self-conjugate subgroup.

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