Let $K$ be a normal subgroup of order 2 in group $G$, show that $K$ lies in the centre of $G$. Describe a surjective homomorphism of the orthogonal group $\mathrm{O}(3)$ onto $C_2$ and another onto the special orthogonal group $\mathrm{SO}(3)$.
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Zev Chonoles
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Since $K$ is a normal subgroup of order $2$, there is only one nonidentity element, say $a\in K$. Then for all $g\in G$, we have that $gag^{-1}=e$ or $gag^{-1}=a$. If the latter occurs, then $K\subseteq Z(G)$ and we're done since $(gag^{-1})g=ga=ag$. Assume that $gag^{-1}=e$. Then $ga=g$, thus multiplying on the left by $g^{-1}$, we get $a=e$, a contradiction. Thus, $K\subseteq Z(G)$ as desired.
Clayton
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Hint: Prove that $[x,y]\in K$ for every $y\in G$, where $x$ is the generator of $K$.
Alexander Gruber
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$[x,y]=x^{-1}y^{-1}xy$. +1 – Mikasa Dec 30 '12 at 09:37
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Center of a group Z(G) is the sub group of elements that commute with all members of the group. A subgroup of order two has two elements: identity element and another element, say x, which is self inverse. Since Z(G) is a subgroup it contains the identity element. We show that the other element x also is in Z(G).
Suppose K is the normal subgroup of order 2 and K={1, x}. If g is an arbitrary element of G, then gK = Kg as K is normal. That is {g, gx} = {g, xg}. Since the two sets are equal and x is not identity, hence gx = xg. This implies that x is in the Center of the Group (as it commutes with an arbitrary element of the group)