Math 360, Fall 2013, Assignment 7 Proof 1

Definition: Let G be any group, and let H be a subgroup of G. Define the relation left congruence modulo H, denoted $\sim_{L,H}$. as:
$a \sim_{L,H} b \Leftrightarrow b^{-1}a\in H$

Theorem: $\sim_{L,H}$ is always an equivalence relation

Proof:
To show that $\sim_{L,H}$ is an equivalence relation, we must show that it is reflexive, symmetric, and transitive.

$\sim_{L,H}$ is reflexive:
Choose an $a\in G$ to be arbitrary. Note that $a\sim_{L,H}a$ can only be true if $a^{-1}a\in H$. But we know that $a^{-1}a=1$, by the definition of an inverse, and by the definition of a subgroup, the identity must be contained. Therefore, we conclude that $a\sim_{L,H} a$, and since we chose a to be arbitrary, we conclude that $\sim_{L,H}$ is reflexive.

$\sim_{L,H}$ is symmetric:
Choose two elements, $a,b \in G$, to be arbitrary. Suppose that $a\sim_{L,H} b$. If this is true, then $b^{-1}a \in H$. But since H is a subgroup of G, by the definition of subgroup, it must be closed under inversion. Therefore, it must also be true that $a^{-1}b\in H$. Note that to calculate the inverse of $b^{-1}a$, the letters are first reversed, and then the exponents are multiplied by $-1$. But if $a^{-1}b\in H$, we conclude $b\sim_{L,H} a$. Since we chose a and b to be arbitrary, we conclude that $\sim_{L,H}$ is symmetric.

$\sim_{L,H}$ is transitive:
Choose three elements, $a,b,c \in G$ to be arbitrary. Suppose both that $a\sim_{L,H} b$ and $b\sim_{L,H} c$. We must show that $a\sim_{L,H} c$. However, we know two things, that $b^{-1}a \in H$ and that $c^{-1}b$ in H. But since H is a subgroup, it must be closed under the operation. Here, we can use multiplicative notation to write $(c^{-1}b)(b^-1a) \in H$. However, since H is a subgroup of G, the operation which is induced upon it must by definition be associative. Therefore, we may change the parenthesis to say $(c^{-1})(b^{-1}ba) \in H$. Now, we recall that by the definition of an inverse element, $b^{-1}b=1$. Therefore, $c^{-1}a\in H$, which means that $a\sim_{L,H}c$. Since we chose a, b and c to be arbitrary, we conclude that $\sim_{L,H}$ is transitive.

Since we have shown that $\sim_{L,H}$ is reflexive, symmetric and transitive, we conclude that $\sim_{L,H}$ is an equivalence relation.

Q.E.D.