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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 $\text{cl}(a)=\{x^{-1}ax\mid x\in G\}$.
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 $G$. We can then count the number of elements of $G$ by instead counting the number of elements in each 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.
- For each $a\in G$, we know that $a$ is a conjugate of $a$.
- If $a$ is a conjugate of $b$, show that $b$ is a conjugate of $a$.
- If $a$ is a conjugate of $b$ and $b$ is a conjugate of $c$, show that $a$ is a conjugate of $c$.
Problem.The number of conjugates of $a$
Let $G$ be a group and let $a\in G$. Then the number of conjugates of $a$ equals the index of the centralizer of $a$ in $G$. In symbols, we write $|\cl(a)|=|G:C(a)|$.
Problem. Second Isomorphism Theorem
Suppose that $K$ is a subgroup of $G$ and $N$ is a normal subgroup of $G$. Prove that $K/(K\cap N)\approx KN/N$.
Problem. Third Isomorphism Theorem
Suppose that $M$ and $N$ are normal subgroups of $G$ with $N\leq M$. Prove that $(G/N)/(M/N)\approx G/M$.
Problem. Butterfly Lemma
Later
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 $G$.
For more problems, see AllProblems