Third Isomorphism Theorem/Groups/Proof 1
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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 $H$.
We define a mapping:
- $\phi: G / N \to G / H$ by $\map \phi {g N} = g H$
Since $\phi$ is defined on cosets, we need to check that $\phi$ is well-defined.
Suppose $x N = y N$.
Then:
- $y^{-1} x \in N$
Then:
- $N \le H \implies y^{-1} x \in H$
and so:
- $x H = y H$
So:
- $\map \phi {x N} = \map \phi {y N}$
and $\phi$ is indeed well-defined.
Now $\phi$ is a homomorphism, from:
\(\ds \map \phi {x N} \map \phi {y N}\) | \(=\) | \(\ds \paren {x H} \paren {y H}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x y H\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {x y N}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {x N y N}\) |
Also, since $N \subseteq H$, it follows that:
- $\order N \le \order H$
So:
- $\order {G / N} \ge \order {G / H}$, indicating $\phi$ is surjective.
So:
\(\ds \map \ker \phi\) | \(=\) | \(\ds \set {g N \in G / N: \map \phi {g N} = e_{G / H} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {g N \in G / N: g H = H}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {g N \in G / N: g \in H}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds H / N\) |
The result follows from the First Isomorphism Theorem.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Group Homomorphism and Isomorphism: $\S 68$. The First Isomorphism Theorem
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $8$: The Homomorphism Theorem: Theorem $8.16$