91Ó°ÊÓ

A subgroup \(H\) of a group \(G\) is said to be normal if for every element \(h\) in \(H\) and every element \(g\) in \(G\), \(g^{-1}hg\) is also in \(H\). In other words, \(H\) is normal in \(G\) if \(gHg^{-1} = H\) for all \(g\) in \(G\).

We're given that \(G\) is a group with exactly one subgroup \(H\) of order \(k\). We want to prove that \(H\) is normal in \(G\).

Let \(h\) be an element in the subgroup \(H\) and \(g\) be an element from the group \(G\). We want to show \(g^{-1}hg\) are also in \(H\). First, observe that the set \(\{g^{-1}hg: h \in H\}\) is a subgroup of \(G\). Let's call this subgroup \(K\). We can use the elementwise definition of a subgroup to verify this: 1. Identity element: \(g^{-1}eg = g^{-1}g = e\in K\), where "e" is the identity element in \(G\). 2. Inverses: \((g^{-1}hg)^{-1} = (g^{-1}h^{-1}(g^{-1})^{-1})= g^{-1}h^{-1}g\in K\). 3. Closure: Consider \(g^{-1}h_1g\) and \(g^{-1}h_2g\) for \(h_1, h_2 \in H\). Their product is \((g^{-1}h_1g)(g^{-1}h_2g) = g^{-1}(h_1h_2)g\). Since \(h_1h_2\in H\), this product is in the form of \(g^{-1}hg \in K\). Now we know that \(K\) is a subgroup. Additionally, we have that \( |H| = |K| = k\) since the conjugation operation is a bijection on \(H\) (it is one-to-one and onto).

Since the order of \(K\) is the same as the order of \(H\), and there is exactly one subgroup of order \(k\) in \(G\), we conclude that \(K=H\). Thus, for every element \(h\) in \(H\) and every element \(g\) in \(G\), \(g^{-1}hg\) is in \(H\). This means that \(H\) is normal in \(G\).