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A Kick in the Discovery

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Normal Subgroups

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*It is assumed that the reader is familiar with mathematical Groups.*

Imagine this: You’re in the middle of a lecture in your abstract algebra course discussing group theory and all of a sudden you hear that a subgroup is normal. What’s so normal about it?

Unfortunately, I’m not sure why the word normal was chosen, but we can at least learn about them.

Nebulous Normal Subgroups

We are going to build off of the definition of a group. So, recall, the definition of a group:

Definition: (Group) A set G with binary operation \cdot is called a group if the following properties hold: (note some authors have (G0) built into the definition of a binary operation)

(G0). For any a,b\in G, the product a\cdot b \in G,

(G1). The operation \cdot is associative so a\cdot (b\cdot c) = (a\cdot b)\cdot c for all a,b,c\in G,

(G2). There is an identity element e\in G, where a\cdot e = e \cdot a = e for all a\in G, and

(G3). For every a\in G, there is an element a^{-1} \in G, called the inverse of a, such that a\cdot a^{-1} = a^{-1}\cdot a = e.

Whenever you have some sort of structure, it can be interesting to look at substructures (structures within the structure). In this case, a group within a group! Wow, amazing! Guess what we call these structures… we call them subgroups. Clever name huh?

A subgroup is a subset of elements from a group that satisfies (G0)-(G3) in their own right (with the same operation as G). So, subgroups are groups! When we are checking to see if some subset H\subseteq G is a subgroup of G, we already know that G is a group, so we already know that the operation was associative i.e. we don’t need to check (G1). The trouble comes in with (G0), (G2), and (G3).

Since we are taking a subset (call it H) of elements from G, there is no reason why the product of any two of the elements in H must also be in H. For example, consider the group of integers \mathbb{Z} and the operation addition, say you took the subset H=\{1,\;2,\;3\} \subseteq G=\mathbb{Z}. Clearly 1+3 = 4 and not in H. Moreover, the additive identity 0 is not in H, nor are the additive inverses of 1, 2, or 3. Needless to say, H is not close to being a subgroup of the integers!

To prove that H is a subgroup of G we need to show that H contains the identity of G (G2), is closed under the group operation (G0), and is closed with respect to inverses (G3). (NOTE: if you show (G0) and (G3) then these imply (G2). Can you see why?)

If we add one more quality to H we reach what are deemed Normal Subgroups of G.

Definition: (Normal Subgroup) A subgroup N of a group G is called normal if it is closed with respect to conjugates:1

(N1) For all a\in N and for all x\in G it must be that xax^{-1}\in N.

Now, we would need to check (G0), (G2), (G3), and (N1) to prove that N is a normal subgroup. Let’s get some practice with normal subgroups!

Neat Normal Subgroups

Normal Number 1

A really useful property of abelian groups (groups where the operation is commutative) is that every subgroup is normal. The proof is surprisingly quick!

Proof: Let G be an abelian group and let N be a subgroup of G. Since N is a subgroup, we know (G0), (G2), and (G3) are satisfied. Thus, we only need to check (N1)!

Consider some a\in N and x\in G. Then, xax^{-1} = a (xx^{-1}) = ae = a \in N. Where we used that the operation was associative and commutative!

\square

Normal Number 2

If N is a subgroup of G, then N is normal if and only if N has the property: For all a,b\in G where ab \in H iff ba \in H.

This one is just downright hard to parse!

What it’s saying is (i) if N is normal then it has the property: if ab\in N then ba \in N likewise if ba \in N then ab\in N. And, (ii) if N has the property: if ab\in N then ba \in N if and only if ba \in N then ab\in N. Then N is normal.

Proof: So, let’s begin showing the forward direction:

Forward: If N is normal, then it has the property: For all a,b\in G we have ab\in N iff ba \in N.

Let N be a normal subgroup of G. Consider a,b\in G. If ab\in N, then since N is normal and b\in G, we know that b(ab)b^{-1} = (ba)e = ba \in N. Likewise, if ba \in N then a(ba)a^{-1} = (ab)e = ab \in N since a\in G and N is normal.

Backward: If N has the property: For all a,b\in G we have ab\in N iff ba \in N. Then N is normal.

Ok, now let N be a subgroup of G where ab\in N iff ba \in N. We aim to show that N is closed with respect to conjugates ie (N1). Let, a\in N and x\in G. We need to show that xax^{-1}\in N. Indeed since,

a = a(x^{-1}x) = (ax^{-1}) x \in N implies x(ax^{-1}) = xax^{-1}\in N.

\square

Normal Number 3

There are alternative definitions of a normal subgroup. One comes from the following equivalence statement:

A subgroup N of G is normal if and only if gN = Ng for all g\in G. Where, the sets gN and Ng are defined to be:

gN := \{gn \;:\; n\in N\}

Ng := \{ng \;:\; n\in N\}.

*****Unrelated but important. These two sets gN and Ng are called cosets of N*****

Proof: Let N be a subgroup of G. We must prove two implications, let’s break them up.

Forward: If N is a normal subgroup of G, then gN = Ng for all g\in G.

Let N be a normal subgroup of G. Consider some a\in gN. By definition, a = gn for some n\in N. We aim to show that a \in Ng by showing that a is also of the form a = n_2 g for some n_2\in N. Since g\in G we deduce,

a = gn = (gn)(g^{-1}g) = \underbrace{(gng^{-1})}_{=\;n_2\;\in N}g \in Ng.

Where we used that N is normal to conclude that gng^{-1}\in N by property (N1).

Backward: If gN = Ng for all g\in G, then N is normal.

Let gN = Ng for all g\in G. Consider some a\in N and x\in G. We aim to show, xax^{-1}\in N. Indeed, since xa \in xN implies xa \in Nx which itself implies xa = n_ax for some n_a \in N. Therefore,

xa = n_a x \; \implies xax^{-1} = n_a xx^{-1} = n_a e = n_a \in N.

Concluding the proof!

\square

Normal Number 4 (The Finale)

The intersection of two normal subgroups is normal!

Remark: We left this one for last because we must first show that N\cup K is a subgroup. This makes the proof longer than the others.

Proof: Let N and K be two normal subgroups of G. We aim to show N\cup K is a normal subgroup of G. Recall, this means we must show that N\cup K satisfies (G0), (G2), (G3), and (N1). Also, recall,

N\cup K:=\{g\in G \;:\; g\in N\mathrm{and}\;g\in K\}.

(G2). Since N and K be two normal subgroups of G we know that e\in N and e\in K. Thus, e is in both N and K or e\in N\cup K.

(G0) Let a,b\in N\cup K. Then a is in both N and K and b is in both N and K. Consider their product ab. Since N is a subgroup and both a and b are in N, we know that ab is in N since it’s closed under the operation. Likewise with K. Thus, ab\in N\cup K.

(G3) Let a\in N\cup K. Then a is in both N and K again. And again, since N and K are subgroups, we know that a^{-1} is in both N and K. Therefore, a^{-1} \in N\cup K.

(N1) Let a\in N\cup K as well as x\in G. Our goal is to show that xax^{-1}\in N\cup K. Indeed, since xax^{-1}\in N because N is normal and xax^{-1}\in K since K is normal. Thus, xax^{-1} is in both N and K, or xax^{-1}\in N\cup K.

\square

Final Remarks

We have had lots of practice with normal subgroups now. The reason for this is twofold. One, they’re awesome. Two, we want to study and classify the properties that groups have. A useful tool to do so is called a quotient group, and before we can even mention what these are we need to know how to work with normal subgroups!

I leave here two challenges to you, the reader:

(1) Let the center of a group be defined as the set C := \{a\in G\;:\;ax=xa\;\mathrm{for}\;\mathrm{all}\; x\in G\}. Show that C is a normal subgroup! Hint check our Normal Number 1.

(2) Let N be a normal subgroup and K be a subgroup, then the set NK := \{nk\;:\; n\in N\;k\in K\} is a subgroup!

Footnote:

  1. A conjugate of a\in G is any element of the form xax^{-1} where x\in G. ↩︎

2 responses to “Normal Subgroups”

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    […] that it’s assumed the reader is familiar with groups and subgroups. And note to subscribers, the way that the math (or maths for my British friends) is formatted on […]

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    […] that it’s assumed the reader is familiar with groups and subgroups. […]

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