Let \(G\) be a group of order \(10\) with a normal subgroup of order \(2\). Prove that \(G\) is abelian.
Solution
Let \[ H\trianglelefteq G, \qquad |H|=2. \]
1. The subgroup of order \(5\) is unique and normal.
By Sylow’s theorem, the number \(n_5\) of Sylow \(5\)-subgroups satisfies \[ n_5\equiv1\pmod5 \] and \[ n_5\mid2. \] The only possibility is \[ n_5=1. \] Thus there is a unique subgroup \[ K\le G, \qquad |K|=5, \] and uniqueness makes \(K\) normal in \(G\).
2. The two normal subgroups commute elementwise.
Since \(|H|=2\) and \(|K|=5\), \[ H\cap K=\{e\}. \] For \(h\in H\) and \(k\in K\), the commutator \[ [h,k]=hkh^{-1}k^{-1} \] belongs to \(H\) because \(H\) is normal, and belongs to \(K\) because \(K\) is normal. Hence \[ [h,k]\in H\cap K=\{e\}. \] Thus every element of \(H\) commutes with every element of \(K\).
3. The subgroups generate all of \(G\) and are abelian.
Because \(H\cap K=\{e\}\), \[ |HK|=\frac{|H||K|}{|H\cap K|}=10=|G|. \] Hence \[ G=HK. \] The group \(H\) has prime order \(2\), so it is cyclic; likewise \(K\) has prime order \(5\), so it is cyclic. Thus both are abelian, and step 2 shows they commute with each other. Therefore \[ G\cong H\times K\cong C_2\times C_5, \] which is abelian. Hence \[ \boxed{G\text{ is abelian}.} \]