Problem 11 - Abstract Algebra: Theory and Applications

Question

If a group \(G\) has exactly one subgroup \(H\) of order \(k,\) prove that \(H\) is normal in \(G\).

Step-by-Step Solution

Answer

Question: Prove that if \(G\) is a group with exactly one subgroup \(H\) of order \(k\), then \(H\) is normal in \(G\). Answer: To prove that the subgroup \(H\) is normal in \(G\), we showed that for any element \(g\) in the group \(G\) and any element \(h\) in \(H\), the conjugates \(g^{-1}hg\) are also in \(H\). We observed that the set \(\{g^{-1}hg: h \in H\}\) forms a subgroup \(K\) and found that the order of \(K\) is the same as the order of \(H\), which is \(k\). Since there is exactly one subgroup of order \(k\) in \(G\), we concluded that \(K=H\). Therefore, \(H\) is normal in \(G\).

Step 1: Recall the definition of normal subgroups

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 \(G\).

Step 2: Establish the given information

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 \(G\).

Step 3: Show that the conjugates are in some subgroup of order \(k\)

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).

Step 4: Conclude that \(H\) is normal in \(G\)

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 \(G\).