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Proof

(related to Theorem: First Isomorphism Theorem for Groups)

• By hypothesis, $(G,\ast)$, $(H,\cdot)$ are groups and $f:G\to N$ is a group homomorphism with the kernel $\ker(f).$
• We demonstrate that $(G/\ker(f),\circ)$ is a factor group $(G/\ker(f),\circ).$
  • First of all, $\ker(f)=\{g\in G\mid f(g)=e_H\}$ is a subgroup of $G$ according to the corresponding lemma.
  • Moreover, it is even a normal subgroup of $G,$ according to another previous lemma.
  • By definition of factor groups, $(G/\ker(f),\circ)$ is a factor group with the group operation "$\circ$" defined by $$(a_1 \ker(f))\circ(a_2\ker(f)):=(a_1\ast a_2)\ker(f)$$ for all $a_1,a_2\in G.$
• Moreover, the image $\operatorname{im}(f)=\{f(g)\in H\mid g\in G\}$ is a subgroup of $(H,\cdot)$ according to the corresponding lemma, and in particular a group itself.
• Thus, the algebraic structures $(\operatorname{im}(f),\cdot)$ and $(G/\ker(f),\circ)$ are groups.
• The function $\phi:G/\ker(f)\to \operatorname{im}(f),$ $\phi(a\ker(f))=f(a)$ is well-defined:
  • Assume, $a,b\in G$ with $a\ker(f)=b\ker(f).$
  • Therefore $a\ast b^{-1}\in \ker(f).$
  • Since $f$ is a group homomorphism, it follows from the properties of group homomorphism that $$\begin{array}{rcl}f(a\ast b^{-1})&=&f(a)\cdot f(b^{-1})\\&=&f(e_G)\\&=&e_H.\end{array}$$
  • Thus, $f(a)=f(b).$
• Moreover, $\phi$ is itself a group homomorphism.
  • Since for all $a,b\in G$ we have $$\begin{array}{rcl}\phi(a\ker(f)\circ b\ker (f))&=&\phi((a\ast b)\ker(f))\\ &=&f(a\ast b)\\ &=&f(a)\cdot f(b)\\ &=&\phi(a\ker(f))\circ \phi(b\ker(f)).\end{array}$$
• $\phi$ is injective.
  • Let $a\in G$ with $\phi(a\ker(f))=f(a)=e_H$ be given.
  • Then $a\in\ker(f).$
  • Therefore, $a\ker(f)=e_G\ker(f).$
  • It follows $\ker(\phi)=\{e_G\ker(f)\}.$
• $\phi$ is surjective.
  • Let $b\in \operatorname{im}(f).$
  • Then there is an $a\in G$ with $f(a)=b.$
  • It follows $\phi(a\ker(f))=b.$
• Altogether, we have shown that $\phi$ is a bijective group homomorphism, in other words $\phi$ is an isomorphism.
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References

Bibliography

1. Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013