91Ó°ÊÓ

A subgroup \(H\) of a group \(G\) is called a normal subgroup if for every element \(g \in G\) and \(h \in H\), the following condition holds: \(g h g^{-1} \in H\).

To show that the intersection of \(H_1\) and \(H_2\) is a normal subgroup, we first need to show that it is a subgroup. Let's consider two elements, \(h_1,h_2 \in H_1 \cap H_2\). By definition, we know that \(h_1,h_2 \in H_1\) and \(h_1,h_2 \in H_2\). Now, notice that the product of two elements in \(H_1 \cap H_2\) is also in \(H_1 \cap H_2\): - Since \(h_1,h_2 \in H_1\), we have \(h_1 h_2 \in H_1\) (as \(H_1\) is a subgroup). - Similarly, since \(h_1,h_2 \in H_2\), we have \(h_1 h_2 \in H_2\) (as \(H_2\) is a subgroup). - Therefore, \(h_1 h_2 \in H_1 \cap H_2\). Furthermore, the inverse of an element in \(H_1 \cap H_2\) also belongs to the intersection: - If \(h \in H_1 \cap H_2\), then \(h \in H_1\) and \(h \in H_2\). - Since \(H_1\) and \(H_2\) are both subgroups, this implies that \(h^{-1} \in H_1\) and \(h^{-1} \in H_2\). - Therefore, \(h^{-1} \in H_1 \cap H_2\). Hence, we have shown that \(H_1 \cap H_2\) is a subgroup of \(G\).

Now, we need to verify whether \(H_1 \cap H_2\) is a normal subgroup of \(G\). To do this, we will check the normal subgroup condition mentioned in Step 1: Let \(g \in G\) and \(h \in H_1 \cap H_2\). We need to show that \(g h g^{-1} \in H_1 \cap H_2\). Since \(h \in H_1 \cap H_2\), it is also true that \(h \in H_1\) and \(h \in H_2\). Since \(H_1\) and \(H_2\) are both normal subgroups, this implies that: - \(g h g^{-1} \in H_1\), because \(H_1\) is normal and \(h \in H_1\), - \(g h g^{-1} \in H_2\), because \(H_2\) is normal and \(h \in H_2\). So, \(g h g^{-1}\) is in both \(H_1\) and \(H_2\), which means \(g h g^{-1} \in H_1 \cap H_2\), proving that \(H_1 \cap H_2\) satisfies the normal subgroup condition.

Therefore, we have shown that the intersection of two normal subgroups, \(H_1\) and \(H_2\), of a group \(G\) is also a normal subgroup of \(G\).