normal subgroup

Normal subgroup

For a group $(G,\circ )$ and subgroup $U\le G$, the following are equivalent:

1. $U$ is normal: $U⊴G$
2. $\forall a\in G:a^{-1}\circ U\circ a\subseteq U$ ($U$ is closed under conjugation)
3. $\forall a\in G$: $a\circ U=U\circ a$ (left and right cosets form the same partition, i.e. are equal)

For a commutative group, e.g. abelian groups, this is trivially satisfied for any subgroup → all subgroups are normal.

Proof: We show $(2)⟺(3)$: $x\in U⟹a^{-1}\circ x\circ a=y\in U$ “$\Rightarrow$”: (“von links”) $x\circ a=a\circ y$ $x\circ a\in U\circ a\underset{(U⊴G)}{\underbrace{=a\circ U}}⟹\exists z\in U:x\circ a=a\circ z⟹a\circ z=a\circ y⟹z=y\in U$ “$\Leftarrow$”: (“von rechts”) $\forall a\in G:a^{-1}\circ U\circ a\subseteq U,x\in U⟹a\circ x\circ a^{-1}\in U$ (Voraussetzung) $⟹x\circ a\in a\circ U$ $⟹U\circ a\subseteq a\circ U$ And $a\circ x\circ a^{-1}$ $⟹a\circ x\in U\circ a$ $⟹a\circ U\subseteq U\circ a$ → $U⊴G$

When you "view" the subgroup from different perspectives in the group (by conjugating with different elements), you always see the same set of elements.

The trivial subgroup $\{e\}$ is always normal.

For any $a\in G$: $a^{-1}ea=e$ since $e$ is the identity. This means:

1. $a^{-1}\{e\}a=\{e\}$ for all $a\in G$, satisfying the conjugation definition of normality
2. $a\{e\}=\{e\}a=\{a\}$ for all $a\in G$, so left and right cosets are always equal

Index-2 Subgroup

Let $G$ be a group and $U\le G$ a subgroup with index 2 ($[G:U]$), meaning there are exactly two distinct cosets of $U$ in $G$. Then these cosets are:

1. $U$ itself (since $eU=U$ where $e$ is the identity)

2. The set difference $G\setminus U$ (since together they form $G$ but are disjoint (since they form a partition))

Consider the group $\mathbb{Z}$ under addition and its subgroup $2\mathbb{Z}$ (even integers): The two cosets are:

• $2\mathbb{Z}=\{\dots ,-4,-2,0,2,4,\dots \}$ (the subgroup itself)
• $1+2\mathbb{Z}=\{\dots ,-3,-1,1,3,\dots \}$ (the odd integers)

Together these form a partition of $\mathbb{Z}$, and indeed $1+2\mathbb{Z}=\mathbb{Z}\setminus 2\mathbb{Z}$

$(\tilde{a}\circ \tilde{b})\circ N=(a\circ b)\circ NN⊴(G,\circ )$ … generalizing modulo arithmetic …

Different representations of the same coset/element in the factor group.

For the abelian group $(\mathbb{Z},+)$ and $m\ge 1$, its $m$ cosets area normal subgroup $m\mathbb{Z}⊴\mathbb{Z}$.
Looking at the concrete case of ${\mathbb{Z}}_4$, we have the following congruence classes:
$0+4\mathbb{Z}=\bar{0}=4\mathbb{Z}$ (note: $\bar{0}$ is the neutral element of the factor group)
$1+4\mathbb{Z}=\bar{1}$
$2+4\mathbb{Z}=\bar{2}$
$3+4\mathbb{Z}=\bar{3}$
We can do arithmetic with these classes, e.g.: $(1+4\mathbb{Z})+(2+4\mathbb{Z})=3+4\mathbb{Z}$
This is the same as the addition in ${\mathbb{Z}}_4$, adding the elements and taking the remainder modulo 4.
Same as above, but a different representation: $(17+4\mathbb{Z})+(38+4\mathbb{Z})=55+4\mathbb{Z}$
The reason we can do this is because $4\mathbb{Z}⊴\mathbb{Z}$.

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For any group $(G,\circ )$ with a normal subgroup $N⊴G$ and the coses $a\circ N,a\in G$ (left and right coset equiv).
Now (like above) we define the same operation on the cosets as on the group, i.e. $(a\circ N)\circ (b\circ N)=(a\circ b)\circ N$
We can now take this concept of modulo arithmetic and generalize it it to any group with a normal subgroup.
$\tilde{a}\in a\circ N=\tilde{a}\circ N$, $\tilde{b}\in b\circ N=\tilde{b}\circ N$
$(\tilde{a}\circ N)\circ (\tilde{b}\circ N)=(a\circ b)\circ N$

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Proof: $(\tilde{a}\circ \tilde{b})\circ N=(a\circ b)\circ N$
$\exists n_1,n_2,n_3,n_4\in N:$
$\tilde{a}=a\circ n_1=n_2\circ a$ ($n_1,n_2$ are representatives of the same coset)
$\tilde{b}=b\circ n_3=n_4\circ b$
$\tilde{a}\circ \tilde{b}=a\circ n_1\circ b\circ n_3=\underset{u\in N\circ (a\circ b)=(a\circ b)\circ N}{\underbrace{n_2\circ a\circ b}}\circ n_3=\dots$
$(⟹\exists n_5\in N:u=a\circ b=n_5)$
$\cdots =a\circ b\circ n_5\circ n_3\in (a\circ b)\circ N$ (since $n_5\circ n_3\in N$ and $N$ being a subgroup and hence closed).
$⟹(\tilde{a}\circ \tilde{b})\circ N=(a\circ b)\circ N$ $✓$

Clear up connection to change of basis and conjugation …

Think of conjugation ( $a^{-1}na$) as viewing a subgroup from different "vantage points" within the group:

For a regular subgroup, this view might change depending on where you stand (which element $a$ you conjugate with).
For a normal subgroup, the view remains the same no matter where you stand - it’s “invariant” under these perspective changes.
In mathematical terms, conjugating any element of a normal subgroup always gives you another element that’s still in the subgroup.

When working with matrices, conjugation $A^{-1}BA$ represents a change of basis. A normal subgroup being "invariant" means it looks the same in any basis - it captures some fundamental property that doesn't depend on how we choose to view it.