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\( \newcommand{\cat}[1] {\mathrm{#1}} \newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})} \newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}} \newcommand{\multiBetaReduction}[0] {\twoheadrightarrow_{\beta}} \newcommand{\betaReduction}[0] {\rightarrow_{\beta}} \newcommand{\betaEq}[0] {=_{\beta}} \newcommand{\string}[1] {\texttt{"}\mathtt{#1}\texttt{"}} \newcommand{\symbolq}[1] {\texttt{`}\mathtt{#1}\texttt{'}} \newcommand{\groupMul}[1] { \cdot_{\small{#1}}} \newcommand{\inv}[1] {#1^{-1} } \newcommand{\bm}[1] { \boldsymbol{#1} } \require{physics} \require{ams} \)
Math and science::Algebra::Aluffi

Subgroups. Propositions.

The proofs of the following are on the reverse side. Can you remember them?

Image of subgroup is a subgroup in \( \cat{Grp} \)

Let \( \varphi : G \to H \) be a group homomorphism, and let \( G' \) be a subgroup of \( G \). Then the image of \( G' \) through \( \varphi \) is a subgroup of \( H \).

Let \( \varphi : G \to H \) be a group homomorphism, and let \( H' \) be a subgroup of \( H \). Then [what can be said about \( \inv{\varphi}(H) \)?].

The proof has two steps, one for each of the requisite conditions of a subgroup.

Let \( H_1 \) and \( H_2 \) be two subgroups of \( G \). Then \( H_1 \cap H_2 \) is a subgroup of \( G \).