Dummit And Foote Solutions Chapter 4 Overleaf High Quality <Mobile>

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\beginsolution Let $|H| = n$ and suppose $H$ is the only subgroup of $G$ with order $n$. For any $g \in G$, consider $gHg^-1$. Conjugation is an automorphism of $G$, so $|gHg^-1| = |H| = n$. Thus $gHg^-1$ is also a subgroup of $G$ of order $n$. By uniqueness, $gHg^-1 = H$ for all $g \in G$. Hence $H \trianglelefteq G$. \endsolution Dummit And Foote Solutions Chapter 4 Overleaf High Quality