A subgroup H of a group G is normal if for every g in G, gH=Hg.
Normal subgroups are an important topic and I will try to go through this slowly. Some hopefully obvious observations first.
1. The whole group G is a normal subgroup of G.
2. The identity element by itself is a normal subgroup of G.
3. If G is abelian, then every subgroup of G is normal.
We will only find non-normal subgroups in non-abelian groups, and the only one I have presented so far is D4. Let's look at the Cayley table.
There are several normal subgroups in D4, including the subgroup of all the rotations {I, R90°, R180°, R270°}. In everyday language, if you rotate a rotation, you will get another rotation, and if you rotate a mirror, you will get another mirror.
There are four subgroups of order 2 that are not normal, and each one consists of the identity I and one mirror. Let's look at the group {I, M90°}. Some elements will give us gH = Hg, but all we need is one element to screw up and that proves the group is not normal. On the list below, the elements that don't work are written in red. If you want to check my work, multiplying on the left is taking entries from the column of that element and multiplying in on the right is taking the entries from the row.
I{I, M90°} = {I, M90°}, and {I, M90°}I = {I, M90°}
R90°{I, M90°} = {R90°, M135°}, and {I, M90°}R90° = {R90°, M45°}
R180°{I, M90°} = {R180°, M90°}, and {I, M90°}R180° = {R180°, M0°}
R270°{I, M90°} = {R270°, M45°}, and {I, M90°}R270° = {R90°, M135°}
M0°{I, M90°} = {M0°, R180°}, and {I, M90°}M0° = {M0°, R180°}M45°{I, M90°} = {M45°, R270°}, and {I, M90°}M45° = {M45°, R270°}
M90°{I, M90°} = {M90°, I}, and {I, M90°}M90° = {M90°, I}
M135°{I, M90°} = {M135°, R90°}, and {I, M90°}M135° = {M135°, R90°}
In D4, there are two elements that commute with everything, I and R180°. All the rotations commute with each other, and likewise mirrors commute with mirrors.
In any group G, the set of elements that commute with all other elements is called the center and is denoted as Z(G). The Z comes from the German word for center, zentrum. The center is always a normal subgroup. Proving it is a subgroup is left as an exercise for the reader and the answer will be in the comments tomorrow.
The proof Z(G) must be a normal subgroup.
ReplyDeleteObviously, the elements commute with every other element by definition, but is is a subgroup? We need to prove three things, none of them difficult.
1. The identity is an element of Z(G). It always commutes with every element, so it must be in Z(G).
2. If a is an element of Z(G), we need a^-1 to be there as well.
ax = xa for all x in G.
Multiply the terms on both sides of the equation by a^-1 on both the front and the back.
a^-1axa^-1 = a^-1xaa^-1
Both (a^-1a) and (aa^-1) are the identity, so we can simplify to
xa^-1 = a^-1x
and step 2 is complete.
3. If a and b are elements of Z(G), then ab must be an element.
abx = axb = xab
and we are done.