= First isomorphism theorem
{title2=$G/\ker\varphi\cong\operatorname{im}\varphi$}
= First isomorphism theorem for groups
{synonym}
For a <group homomorphism> $\varphi:G\to H$, the <group isomorphism> $g\ker\varphi\mapsto\varphi(g)$ gives
$$
G/\ker\varphi\cong\operatorname{im}\varphi.
$$
The <group kernel> records which inputs become indistinguishable, and the image records which outputs actually occur. The <first isomorphism theorem> says that identifying precisely those indistinguishable inputs gives a <group> isomorphic to the image. It is one of the general <isomorphism theorems>. https://ocw.mit.edu/courses/18-703-modern-algebra-spring-2013/b40d60898322bbd1d63410b0a925946a_MIT18_703S13_pra_l_10.pdf[McKernan, MIT 18.703, Theorem 10.1: the first isomorphism theorem]{ref}.
For example, reduction modulo three gives a <surjective group homomorphism> from the additive <cyclic group> $\mathbb Z/6\mathbb Z$ to $\mathbb Z/3\mathbb Z$. Its <group kernel> is $\{0,3\}$. The three <cosets> $\{0,3\}$, $\{1,4\}$, and $\{2,5\}$ become the three outputs zero, one, and two, respectively. Thus
$$
(\mathbb Z/6\mathbb Z)/\{0,3\}\cong\mathbb Z/3\mathbb Z.
$$
The <quotient group> diagram shows both the identification and the resulting operation. More generally, <fibres of a group homomorphism are cosets of its kernel> explains why this map is well-defined and <injective>; it is <surjective> onto the image by definition.
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