Groups TT23, First isomorphism theorem

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

Flashcards

@State the first isomorphism theorem.

first-isomorphism-theorem-state

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?

first-isomorphism-theorem-intuition

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 homomorphism. 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?

first-isomorphism-theorem-map
\[g\ker\phi \mapsto \phi(g)\]

@Prove, by appealing to the first isomorphism theorem, that $\mathbb C^\ast / S^1 \cong \mathbb R^\ast$.

quotient-c-star-by-circle-example

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

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

What homomorphism could you define to show

\[\frac{GL _ n(\mathbb F)}{SL _ n(\mathbb F)} \cong \mathbb F^\ast\]

?

quotient-gl-by-sl-example
\[f:GL _ n(\mathbb F) \to \mathbb F^\ast\] \[A \mapsto \det A\]

Given a homomorphism $\phi : G \to H$, what can you say about $ \vert G \vert $?

homomorphism-order-relation
\[ \vert G \vert = \vert \ker \phi \vert \times \vert \text{Im } \phi \vert \]

What’s a general strategy for finding the number of homomorphisms between two groups $G$ and $H$?

counting-homomorphisms-strategy
  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 $\mathrm{Aut}(G/N)$. Then the number of homomorphisms from $G$ to $H$ is given by
\[\sum _ {n(N)>0} n(N) \times \vert \mathrm{Aut}(G/N) \vert \]

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 $g \ker \phi \mapsto \phi(g)$
first-isomorphism-theorem-proof

Todo.

@todo~



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