Groups TT23, First isomorphism theorem

Created: May 04, 2023 | About these notes | View in context | Study these flashcards

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Can you state the first isomorphism theorem?

Let $\phi : G \to H$ be a group homomorphism. Then

  1. $\text{Im } \phi$ is a subgroup of $H$
  2. $\ker \phi$ is a normal subgroup of $G$
  3. $\frac{G}{\ker \phi} \cong \text{Im } \phi$ under the isomorphism $g \ker \phi \mapsto \phi(g)$

The last part of the first isomorphism theorem, that if $\phi : G \to H$ is a group homomorphism then $\frac{G}{\ker \phi} \cong \text{Im } \phi$, seems a little abstract. Can you give an intuitive reason why this is true?

The way in which $\phi$ “collapses” down the group is captured in how it maps multiple elements to the identity element. $\frac{G}{\ker \phi}$ consists of these equivalence classes that are mapped to the same thing by $\phi$. So there is a natural bijection $g \ker \phi \mapsto \phi(g)$.

Let $\phi : G \to H$ be a group homomorpism. Then the first isomorphism theorem states

  1. $\text{Im } \phi$ is a subgroup of $H$
  2. $\ker \phi$ is a normal subgroup of $G$
  3. $\frac{G}{\ker \phi} \cong \text{Im } \phi$

What isomorphism lets you do this?
\[g\ker\phi \mapsto \phi(g)\]

By appealing to the first isomorphism theorem, prove that $C^\ast / S^1 \cong \mathbb R^\ast$?

Note that $\phi : z \mapsto \log \vert z \vert $ has kernel $S^1$, so

\[\begin{aligned} C^\ast / S^1 &= \frac{C^\ast}{\ker \phi} \\\\ &\cong \mathbb R^\ast \end{aligned}\]

What homomorphism could you define to show

\[\frac{GL_n(\mathbb F)}{SL_n(\mathbb F)} \cong \mathbb F^\ast\]

?
\[f:GL_n(\mathbb F) \to \mathbb F^\ast\] \[A \mapsto \det A\]

Given a homomorphism $\phi : G \to H$, what can you say about $H$?
\[|G| = |\ker \phi| \times |\text{Im } \phi|\]

What’s a general strategy for finding the number of homomorphisms between two groups $G$ and $H$?
  1. Determine the normal subgroups $N$ of $G$
  2. Determine $n(N)$, the number of subgroups in $H$ isomorphic to $G/N$
  3. For the normal subgroups where $n(N)>0$, determine the order of $\text{Aut}(G/N)$. Then the number of homomorphisms from $G$ to $H$ is given by
\[\sum_{n(N)>0} n(N) \times |\text{Aut}(G/N)|\]

Proofs

Prove the last part of the first isomorphism theorem:

Let $\phi : G \to H$ be a group homomorphism. Then

  1. $\text{Im } \phi$ is a subgroup of $H$
  2. $\ker \phi$ is a normal subgroup of $G$
  3. $\frac{G}{\ker \phi} \cong \text{Im } \phi$ under the isomorphism $g \ker \phi \mapsto \phi(g)$

:: Todo.



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