Second Isomorphism Theorem/Groups

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Theorem

Let $G$ be a group, and let:

$(1): \quad H$ be a subgroup of $G$
$(2): \quad N$ be a normal subgroup of $G$.


Then:

$\dfrac H {H \cap N} \cong \dfrac {H N} N$

where $\cong$ denotes group isomorphism.


Proof

The fact that $N$ is normal, together with Intersection with Normal Subgroup is Normal, gives us that $N \cap H \lhd H$.

Also, $N \lhd N H = \gen {H, N}$ follows from Subset Product with Normal Subgroup as Generator.


Now we define a mapping $\phi: H \to H N / N$ by the rule:

$\map \phi h = h N$

Note that $N$ need not be a subset of $H$.

Therefore, the coset $h N$ is an element of $H N / N$ rather than of $H / N$.

Then $\phi$ is a homomorphism, as:

$\map \phi {x y} = x y N = \paren {x N} \paren {y N} = \map \phi x \map \phi y$

Then:

\(\ds \map \ker \phi\) \(=\) \(\ds \set {h \in H: \map \phi h = e_{H N / N} }\)
\(\ds \) \(=\) \(\ds \set {h \in H: h N = N}\)
\(\ds \) \(=\) \(\ds \set {h \in H: h \in N}\)
\(\ds \) \(=\) \(\ds H \cap N\)


Then we see that $\phi$ is a surjection because $h n N = h N \in H N / N$ is $\map \phi h$.

The result follows from the First Isomorphism Theorem.

$\blacksquare$


Also known as

This result is also referred to by some sources as the first isomorphism theorem.


Also see


Sources

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