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Normal subgroups of order 2 are central


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.2 Exercise 2.2.10

Let $H$ be a subgroup of order 2 in $G$. Show that $N_G(H) = C_G(H)$. Deduce that if $N_G(H) = G$, then $H \leq Z(G)$.


Solution: Say $H = \{ 1, h \}$.

We already know that $C_G(H) \subseteq N_G(H)$. Now suppose $x \in N_G(H)$; then $$\{ x1x^{-1}, xhx^{-1} \} = \{ 1, h \}.$$ Clearly, then, we have $xhx^{-1} = h$. Thus $x \in C_G(H)$. Hence $N_G(H) = C_G(H)$.

If $N_G(H) = G$, we have $C_G(H) = G$. Then $ghg^{-1} = h$ for all $h \in H$, so that $gh = hg$ for all $h \in H$, and thus $H \leq Z(G)$.


Linearity

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