Proving a group is a normal subgroup from its order

Consider the normalizer $N_G(H)$. We can say that $H\subseteq N_G(H)\subseteq G$.

By Lagrange's theorem, $|H|$ divides $|N_G(H)|$ which also divides $|G|$.

Since $[G:H]=5$ is prime, there are no intermediate divisors between $|H|$ and $|G|$, so we conclude that $|N_G(H)|$ must be one of these, so $N_G(H)$ is one of $H$ or $G$. Can you finish?


I assume $a \notin H$.

If $a^2 \in H \cup aH$ then $H \cup aH$ is a subgroup of $G$. By Lagrange theorem, this is impossible. Similarly, we see that $a^3 \notin H \cup aH \cup a^2H$ and $a^4 \notin H \cup aH \cup a^2H \cup a^3H$. Thus, we have five disjoint cosets of $H$ in $G$: $H$, $aH$, $a^2H$, $a^3H$, $a^4H$. Their union is all of $G$.

Also, $a^kH = Ha^k$ for every $k=0, 1, 2, 3, 4$. Thus, $H$ is normal.

Tags:

Abstract Algebra

Group Theory

Normal Subgroups

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