795: Intersection of Subgroup of Group and Normal Subgroup of Group Is Normal Subgroup of Subgroup

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that intersection of subgroup of group and normal subgroup of group is normal subgroup of subgroup

---

Topics

About: group

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

---

Starting Context

• The reader knows a definition of normal subgroup of group.
• The reader admits the proposition that for any group and its any subgroup, the subgroup is a normal subgroup if its conjugate subgroup by each element of the group is contained in it.

---

Target Context

• The reader will have a description and a proof of the proposition that for any group, the intersection of any subgroup of the group and any normal subgroup of the group is a normal subgroup of the subgroup.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$G$: $\in \{\text{ the groups }\}$
$G_1$: $\in \{\text{ the subgroups of }G\}$
$G_2$: $\in \{\text{ the normal subgroups of }G\}$
//

Statements:
$G_1\cap G_2\in \{\text{ the normal subgroups of }{G}_1\}$
//

---

2: Natural Language Description

For any group, $G$, any subgroup, $G_1\subseteq G$, and any normal subgroup, $G_2\subseteq G$, $G_1\cap G_2\subseteq G_1$ is a normal subgroup of $G_1$.

---

3: Proof

Whole Strategy: Step 1: see that for each $p_1\in G_1$, $p_1(G_1\cap G_2){p_1}^{-1}\subseteq G_1\cap G_2$ and conclude the proposition.

Step 1:

By the proposition that for any group and its any subgroup, the subgroup is a normal subgroup if its conjugate subgroup by each element of the group is contained in it, it suffices to show that for each $p_1\in G_1$, $p_1(G_1\cap G_2){p_1}^{-1}\subseteq G_1\cap G_2$.

$p_1(G_1\cap G_2){p_1}^{-1}\subseteq p_1{G}_2{p_1}^{-1}=G_2$, because $p_1\in G$. $p_1(G_1\cap G_2){p_1}^{-1}\subseteq p_1{G}_1{p_1}^{-1}\subseteq G_1$, because $p_1\in G_1$. So, $p_1(G_1\cap G_2){p_1}^{-1}\subseteq G_1\cap G_2$.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>