Normal Subgroup K in HK

We are given a group \(G\), a subgroup \(H\) of \(G\), and a normal subgroup \(K\) of \(G\). We need to determine the relationship between \(K\) and the product set \(HK = \{hk \mid h \in H, k \in K\}\). The question asks whether \(K\) is a normal subgroup of \(HK\).

Understanding Normal Subgroups

Recall the definition of a normal subgroup. A subgroup \(N\) of a group \(M\) is called a normal subgroup of \(M\) if for every element \(m \in M\) and every element \(n \in N\), the element \(mnm^{-1}\) is in \(N\). This can be written as \(mNm^{-1} \subseteq N\) for all \(m \in M\).

In our case, we want to check if \(K\) is a normal subgroup of \(HK\). This means we need to verify if for every element \(x \in HK\) and every element \(k' \in K\), the element \(xk'x^{-1}\) belongs to \(K\).

Proof that K is a Normal Subgroup of HK

Let \(x\) be an arbitrary element from \(HK\). By the definition of \(HK\), \(x\) can be written in the form \(x = hk\), where \(h \in H\) and \(k \in K\).

We need to show that for any \(k' \in K\), the element \(xk'x^{-1}\) is in \(K\).

Let's find the inverse of \(x = hk\). The inverse is \(x^{-1} = (hk)^{-1} = k^{-1}h^{-1}\).

Now, let's compute \(xk'x^{-1}\): \[ xk'x^{-1} = (hk)k'(k^{-1}h^{-1}) \]

We can rearrange the terms. Using associativity, we have: \[ xk'x^{-1} = h(k k' k^{-1})h^{-1} \]

Since \(K\) is a subgroup of \(G\), and \(k, k', k^{-1}\) are all elements of \(K\), their product \(kk'k^{-1}\) is also an element of \(K\). Let \(k'' = kk'k^{-1}\). So, \(k'' \in K\).

Now the expression becomes: \[ xk'x^{-1} = h k'' h^{-1} \]

We are given that \(K\) is a normal subgroup of \(G\). This means that for any element \(g \in G\) and any element \(k'' \in K\), the element \(gk''g^{-1}\) must belong to \(K\).

Since \(H\) is a subgroup of \(G\), any element \(h \in H\) is also an element of \(G\). So, for \(h \in H \subseteq G\) and \(k'' \in K\), the element \(hk''h^{-1}\) must be in \(K\).

Therefore, \(xk'x^{-1} = hk''h^{-1} \in K\).

This holds for any \(x \in HK\) and any \(k' \in K\). Thus, \(K\) satisfies the condition for being a normal subgroup of \(HK\).

Conclusion on K and HK

Based on the definition of a normal subgroup and the properties of the given group \(G\), subgroup \(H\), and normal subgroup \(K\), we have shown that for any element \(x\) in the product set \(HK\), the conjugate \(xKx^{-1}\) is contained within \(K\). This directly proves that \(K\) is a normal subgroup of \(HK\).

Let's summarize why this works:

• \(K\) is normal in \(G\), meaning \(gKg^{-1} \subseteq K\) for all \(g \in G\).
• Any element \(x \in HK\) is of the form \(hk\) for some \(h \in H\) and \(k \in K\).
• The conjugate \(xKx^{-1} = (hk)K(hk)^{-1} = hkKk^{-1}h^{-1}\).
• Since \(K\) is a subgroup, \(Kk^{-1} = K\). So, the expression simplifies to \(hKh^{-1}\).
• Since \(h \in H \subseteq G\) and \(K\) is normal in \(G\), \(hKh^{-1} \subseteq K\).
• Therefore, \(xKx^{-1} \subseteq K\) for all \(x \in HK\), which means \(K\) is normal in \(HK\).

This result is a fundamental property in group theory when dealing with products of subgroups where one is normal in the larger group.