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My instructor wrote that "[a] quotient group is a group obtained by aggregating similar elements of a larger group using an equivalence relation. In a quotient of a group, the equivalence class of the identity element is always a normal subgroup of the original group, and the other equivalence classes are the cosets of this normal subgroup ...."

He gave an example $\mathbb Z_6 /\{0, 2, 4\}$. The operation is addition mod 6. I was able to write down the equivalence relation $x \equiv 0 \leftrightarrow 3x = 0 \space mod \space 6 $. Indeed, $\{0, 2, 4\}$ is a normal subgroup of $\mathbb Z_6$, and $\{1, 3, 5\}$ is the other coset of $\mathbb Z_6 /\{0, 2, 4\}$.

The question is how do I prove the statement in general, without knowing what the equivalence relation is? I do not think that the reflexive, symmetric, and transitive properties alone are sufficient. For example, if $a \in [\space 0 \space]_\equiv $, how do I know that $-a \in [\space 0 \space]_\equiv $?

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2 Answers 2

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You are right. It also has to be compatible with the operations: the group operation and the inverse, meaning that $\equiv$ is also required to satisfy $$a\equiv a',\ \ \ b\equiv b'\ \implies\ ab\equiv a'b',\ \ \ a^{-1}\equiv(a')^{-1}\,. $$

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Any such equivalence must be assumed to arise from a partition that respects the group operation, meaning to compute the product of $A$ and $B$ (sets in the partition), you can select $a \in A$ and $b \in B$ and the product $AB$ will be the partition containing $ab$. If the partitions form a group under this operation then the partition containing 1 will be a normal subgroup of the original group, and the other sets in the partition will be cosets of it. (prove this)

If your partition starts out as cosets of a normal subgroup to begin with, as in your quoted passage, then there is nothing to check. It's just a restatement of what it means to form the quotient group.

Work more examples.

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