596: Subgroup of Group Multiplied by Normal Subgroup of Group Is Subgroup of Group

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

description/proof of that subgroup of group multiplied by normal subgroup of group is subgroup of group

---

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.

---

Target Context

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

---

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:
$G_1{G}_2\in \{\text{ the subgroups of }G\}$
$\wedge$
$G_2{G}_1\in \{\text{ the subgroups of }G\}$
//

---

2: Natural Language Description

For any group, $G$, any subgroup, $G_1$, of $G$, and any normal subgroup, $G_2$, of $G$, $G_1{G}_2$ is a subgroup of $G$ and $G_2{G}_1$ is a subgroup of $G$.

---

3: Proof

1st let us think of $G_1{G}_2$.

For the identity element, $1\in G$, $1\in G_1$ and $1\in G_2$, so, $1\in G_1{G}_2$.

For any $p_1,p_1^{\prime }\in G_1$ and any $p_2,p_2^{\prime }\in G_2$, $p_1{p}_2{p}_1^{\prime }{p}_2^{\prime }=p_1{p}_1^{\prime }{p}_2^{″}{p_1^{\prime }}^{-1}{p}_1^{\prime }{p}_2^{\prime }$, where $p_2^{″}\in G_2$, because as $p_1^{\prime }{G}_2{p_1^{\prime }}^{-1}=G_2$, there is such a $p_2^{″}$; $=p_1{p}_1^{\prime }{p}_2^{″}{p}_2^{\prime }\in G_1{G}_2$.

For any $p_1\in G_1$ and any $p_2\in G_2$, $(p_1{p}_2{)}^{-1}={p_2}^{-1}{p_1}^{-1}={p_1}^{-1}{p}_2^{\prime }{p}_1{p_1}^{-1}$, where $p_2^{\prime }\in G_2$, because as ${p_1}^{-1}{G}_2{p}_1=G_2$, there is such a $p_2^{\prime }$; $={p_1}^{-1}{p}_2^{\prime }\in G_1{G}_2$.

The associativity of multiplications holds, because it holds in the ambient $G$.

Let us think of $G_2{G}_1$.

For the identity element, $1\in G$, $1\in G_1$ and $1\in G_2$, so, $1\in G_2{G}_1$.

For any $p_1,p_1^{\prime }\in G_1$ and any $p_2,p_2^{\prime }\in G_2$, $p_2{p}_1{p}_2^{\prime }{p}_1^{\prime }=p_2{p}_1{p_1}^{-1}{p}_2^{″}{p}_1{p}_1^{\prime }$, where $p_2^{″}\in G_2$, because as ${p_1}^{-1}{G}_2{p}_1=G_2$, there is such a $p_2^{″}$; $=p_2{p}_2^{″}{p}_1{p}_1^{\prime }\in G_2{G}_1$.

For any $p_1\in G_1$ and any $p_2\in G_2$, $(p_2{p}_1{)}^{-1}={p_1}^{-1}{p_2}^{-1}={p_1}^{-1}{p}_1{p}_2^{\prime }{p_1}^{-1}$, where $p_2^{\prime }\in G_2$, because as $p_1{G}_2{p_1}^{-1}=G_2$, there is such a $p_2^{\prime }$; $=p_2^{\prime }{p_1}^{-1}\in G_2{G}_1$.

The associativity of multiplications holds, because it holds in the ambient $G$.

---

References

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