20131209

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To prove that $H$ is a subgroup of $G$, we have been using two different options, namely The Subgroup Test and The Finite Subgroup Test. Our version of the subgroup test requires that you show (1) that $H$ is nonempty, (2) that $H$ is closed under the operation, and (3) that $H$ is closed under taking inverse. There is another test that can often result in shorter proofs, though not always easier to use. It's called the "One-Step Subgroup Test" and requires you show that $H$ is nonempty and that $ab^{-1}\in H$ whenever $a$ and $b$ are in $H$.

Problem (One Step Subgroup Test)

Suppose that $H$ is a nonempty subset of a group $G$. Suppose that $ab^{-1}\in H$ whenever $a,b\in H$. Prove that $H$ is a subgroup of $G$.

Many of the ideas we'll be doing today hinge up on the notation of an equivalence relation. It's way of grouping together common objects. Given a huge collection of things, we like to sort them into piles of common things. Each object gets put in exactly one pile, and no piles overlap. This is what we call a partition. We say that things in the same pile are equivalent. Many of you have seen the following definition in your past.

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Definition (Equivalence Relation)

Let $S$ be set. Let $\cong$ be a relation on $S$, meaning $\cong$ is a collection $\mathscr{C}$ of ordered pairs of $S$. We way that $A\cong B$ if and only if the ordered pair $(A,B)$ is an element of $\mathscr{C}$. We say that $\cong$ is an equivalence relation if and only if

1. (Reflexive) For every $A\in S$, we know that $A\cong A$ (so the ordered pair $(A,A)$ is always in $\mathscr{C}$).
2. (Symmetric) If $A\cong B$, then $B\cong A.$
3. (Transitive) If $A\cong B$ and $B\cong C$, then $A\cong C$.

When we created cosets of a subgroup $H$, we created an equivalence relation on the elements in a group by saying that two elements in the group are "equivalent" if and only if they are in the same coset. We know that this relation is reflexive because for each $a\in G$, we know that $a$ and $a$ are in the same coset, namely the coset $Ha$. We also know that if $a$ and $b$ are in the same coset, then $b$ and $a$ are in the same coset. Finally, we know that if $a$ and $b$ are in the same coset, and $b$ and $c$ are in the same coset, then $a$ and $c$ must be in the same coset. This partitions the group up into groups of "equivalent" elements. We created the factor group of $G$ by $H$ by just using the set product on these cosets.

The same principle applies to comparing groups. We have way of determining when two groups are "the same". We call the groups isomorphic. The next problem has you show that the notation of isomorphisms creates an equivalence relation on the set of all groups.

Problem (Isomorphisms Yield An Equivalence Relation On The Set Of All Groups)

We say that $A$ is isomorphic to $B$ and write $A\approx B$ if and only if there exists a bijection from $A$ to $B$. In this problem, you'll prove that $\approx$ is an equivalence relation on the set of all groups.

1. Let $A$ be a group. Show that $A$ is isomorphic to $A$ by building an isomorphism from $A$ to $A$.
2. Suppose that $A$ is isomorphic to $B$. Prove that $B$ is isomorphic to $A$.
3. Suppose that $A$ is isomorphic to $B$ and suppose that $B$ is isomorphic to $C$. Use this to produce an isomorphism from $A$ to $C$.

We'll now be preparing ourselves to prove the first Sylow theorem. When we study normal subgroups, we look at the product $x^{-1}ax$ quite often. You also studied this idea in linear algebra when you were studying changing bases. We called two matrices $A$ and $B$ similar in linear algebra if $B=P^{-1}AP$ for some invertible matrix $P$. When two matrices were similar, they represented the exact same linear transformation, just using a different basis. One of the biggest ideas in linear algebra is that if you use the eigenvectors as your basis vectors (the columns of $P$), then the matrix $A$ is similar to a diagonal matrix. Studying the products $P^{-1}AP$ had a huge payoff in linear algebra. It does in group theory as well.

Definition (Conjugacy Class Of A)

Let $G$ be a group. Suppose that $a$ and $b$ are elements of $G$. We say that $a$ and $b$ are conjugate in $G$ if $b=x^{-1}ax$ for some $x\in G$, and we say that $b$ is a conjugate of $a$. The set of conjugates of $a$ is called the conjugacy class of $a$ which we denote by $\text{cl}(a)=\{x^{-1}ax\mid x\in G\}$.

Exercise (Conjugacy Classes Of The Automorphisms Of The Square)

Let $G = D_8$, the automorphisms of the square. Compute the conjugacy classes of $G$.

You've seen conjugacy classes before when you looked at similar matrices in linear algebra. If two matrices are similar, then lots of things are known about the matrices, in particular that they have the same eigenvalues, and the same dimensions of eigenspaces. There's an entire world of facts to explore about similar matrices, which is a specific example of a conjugacy class.

For now, let's show that conjugacy classes partition the group $G$. As a side note, every idea we learn here about conjugacy classes will immediately apply to matrix groups. After we have shown that conjugacy classes partition the group $G$, we can then count the number of elements of $G$ by instead counting the number of elements in each distinct conjugacy class. This means we have $G = \bigcup \cl(a)$ and hence $|G|=\sum |\text{cl}(a)|$ where the sum only includes each conjugacy class once.

Problem (Conjugacy Is An Equivalence Relation)

Show that conjugacy is an equivalence relation on $G$. In other words, show the following three things.

1. For each $a\in G$, we know that $a$ is a conjugate of $a$.
2. If $a$ is a conjugate of $b$, show that $b$ is a conjugate of $a$.
3. If $a$ is a conjugate of $b$ and $b$ is a conjugate of $c$, show that $a$ is a conjugate of $c$.

Let's now connect conjugacy to centralizers. The center of a group is the set of all $x$ that commute with every element in the groups. The centralize of an element is the collection of all $x$ that commute with $a$.

Definition (Centralizer Of An Element)

Let $a\in G$. The centralizer of $a$ is the collection of all $x\in G$ such that $ax=xa$, or alternately $x^{-1}ax=a$. We write this set symbolically as $$C(a)=\{x\in G\mid ax=xa\} = \{x\in G\mid x^{-1}ax=a\} .$$

Exercise (Centralizers Of The Automorphisms Of The Square)

Let $G = D_8$, the automorphisms of the square. Compute the centralizes $C(a)$ for each $a\in G$.

If an element is in the centralizer, then we know that we can use that element to show that $a$ is a conjugate of itself. On your midterm you showed that the centralizer of $a$ is a subgroup of $G$. The next problem has you connect centralizers and conjugacy classes.

Problem (The Number Of Conjugates Of A)

Let $G$ be a group and let $a\in G$. Prove that the number of conjugates of $a$ equals the index of the centralizer of $a$ in $G$. In symbols, we write this as $|\text{cl}(a)|=|G:C(a)|$.

For more problems, see AllProblems