Third Isomorphism Theorem/Groups/Proof 3
Theorem
Let $G$ be a group, and let:
- $H, N$ be normal subgroups of $G$
- $N$ be a subset of $H$.
Then:
- $(1): \quad N$ is a normal subgroup of $H$
- $(2): \quad H / N$ is a normal subgroup of $G / N$
- where $H / N$ denotes the quotient group of $H$ by $N$
- $(3): \quad \dfrac {G / N} {H / N} \cong \dfrac G H$
- where $\cong$ denotes group isomorphism.
Proof
From Normal Subgroup which is Subset of Normal Subgroup is Normal in Subgroup, $N$ is a normal subgroup of $H$.
Let $q_H$ denote the quotient mapping from $G$ to $\dfrac G H$.
Let $q_N$ denote the quotient mapping from $G$ to $\dfrac G N$.
Let $\RR$ be the congruence relation defined by $N$ in $G$.
Let $\TT$ be the congruence relation defined by $H$ in $G$.
Thus from Congruence Relation induces Normal Subgroup:
- $q_H = q_\TT$
and:
- $q_N = q_\RR$
where $q_\RR$ and $q_\TT$ denote the quotient epimorphisms induced by $\RR$ and $\TT$ respectively.
We have that:
\(\ds \tuple {x, y}\) | \(\in\) | \(\ds \RR\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x N\) | \(=\) | \(\ds y N\) | Definition of Congruence Modulo Subgroup | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x y^{-1}\) | \(\in\) | \(\ds N\) | Definition of Normal Subgroup | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x y^{-1}\) | \(\in\) | \(\ds H\) | as $N \subseteq H$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x H\) | \(=\) | \(\ds y H\) | Definition of Normal Subgroup | ||||||||||
\(\ds \tuple {x, y}\) | \(\in\) | \(\ds \TT\) | Definition of Congruence Modulo Subgroup |
That is:
- $\RR \subseteq \TT$
Let $\SS$ be the relation on the quotient group $G / N$ which satisfies:
- $\forall X, Y \in G / N: X \mathrel \SS Y \iff \exists x \in X, y \in Y: x \mathrel \TT y$
That is, by definition of $\TT$:
- $\forall X, Y \in G / N: X \mathrel \SS Y \iff \exists x \in X, y \in Y: x H = y H$
Then by Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence: Corollary:
- $\SS$ is a congruence relation on $G / N$.
Hence by Congruence Relation on Group induces Normal Subgroup:
- $\SS$ induces a normal subgroup of $G / N$.
This needs considerable tedious hard slog to complete it. In particular: We need to identify $q_\SS$ with $q_{H / N}$, that is, show that the normal subgroup defined above is $H / N$. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Again from Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence: Corollary:
- there exists a unique isomorphism $\phi$ from $G / N$ to $G / H$ which satisfies:
- $\phi \circ q_\SS \circ q_\RR = q_\TT$
- where $q_\SS$, $q_\RR$ and $q_\TT$ denote the quotient epimorphisms as appropriate.
That is:
- $\phi \circ q_{H / N} \circ q_N = q_H$
and the result follows.
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.19 \ \text {(b)}$