Group epimorphism

Let $𝜑 : 𝐺 \to 𝐻$ be a homomorphism. The following two statements are equivalent¹: group

$𝜑$ is surjective
$𝜑$ is epic in Category of groups

Proof

Since all surjective functions are epic, it follows that surjective homomorphisms are too. The converse requires more care. Let $𝜙 : 𝐺 ↠ 𝐻$ be an epimorphism and $𝐴 = 𝜙 (𝐺)$ . It is sufficient to show $𝐴 = 𝐻$ . Consider the set $𝐺 / 𝐴$ of left-cosets of $𝐴$ in $𝐺$ , and let $𝑢$ be a unique symbol, $𝑆 = 𝐺 / 𝐴 ⨿ {𝑢}$ , and $𝑆!$ denote the permutation group of the set $𝑆$ . We define a homomorphism $𝜓 1 : 𝐻 \to 𝑆!$ so that for any $ℎ, ℎ 1 \in 𝐻$
$𝜓 1 (ℎ) : 𝐴 ℎ 1 \mapsto 𝐴 ℎ 1 ℎ 𝑢 \mapsto 𝑢$
Let $𝜎 \in 𝑆!$ be the permutation which swaps $𝐴$ and $𝑢$ and leaves everything else invariant, and let $ˆ 𝜎$ be its induced inner automorphism. Then $𝜓 2 = ˆ 𝜎 𝜓 1$ is also a homomorphism. Now, for any $𝑎 \in 𝐴$ it follows $𝜓 1 (𝑎)$ leaves both $𝐴$ and $𝑢$ fixed (the latter is by construction), and hence $𝜎$ commutes with $𝜓 1$ and thus $𝜓 2 (𝑎) = ˆ 𝜎 𝜓 1 (𝑎) = 𝜓 1 (𝑎)$ . Hence $𝜓 1 𝜙 = 𝜓 2 𝜙$ , and since $𝜙$ is epic $𝜓 1 = 𝜓 2$ . But thence follows that $𝜎$ commutes with $𝜓 1 (ℎ)$ for all $ℎ \in 𝐻$ , wherefore $𝜓 1 (ℎ)$ must leave $𝐴$ fixed, and thus $𝜓 1 (ℎ) = ℎ 𝐴 = 𝐴$ and thus $ℎ \in 𝐴$ for all $ℎ \in 𝐻$ .

Corollaries

The above argument holds for Category of finite groups, since $𝑆!$ is finite if $𝐺$ and $𝐻$ are finite.

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1970. A Group Epimorphism is Surjective ↩

Table of Contents

Group epimorphism

Corollaries

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Table of Contents

Group epimorphism

Corollaries

Footnotes

Graph View

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