# Significance of Invariant (So Normal) Subgroups and Conjugacy [closed]

I know that there is an equivalence relation on the conjugacy of subgroups of $G$, so they partition $G$. But I keep reading stuff about orbits, stabilizers, conjugacy classes, auto-morphism groups...What is the connection? I understand this is tedious I apologize

Question 1: What significance do Invariant Subgroups have under automorphisms?

Question 2: What does it mean for two subgroups to be a conjugate of one another.

EDIT: Deleted a Question as its premise was wrong.

## closed as unclear what you're asking by Dietrich Burde, freakish, Namaste, Parcly Taxel, Claude LeiboviciDec 8 '17 at 6:10

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• 1: Important for quotient groups $G/N$. 2: $H=gKg^{-1}$, see here. 3: Class equation, etc., see lecture notes of group theory. – Dietrich Burde Dec 7 '17 at 19:13
• 1) I should've mentioned under automorphism 2) After looking what you linked me to, here is a follow-up: can two conjugate subgroups have an empty intersection? which leads me to question 3) Class Equation means nothing to me as i dont know what a centraliser is. Quotient Group's elements are cosets ,if there is an equivalence relation between each conjugate SUBGROUP, then shouldn't they partition $G$? If they do, then how is that helpful to know? – Kam Dec 7 '17 at 19:24
• @Kam No two subgroups intersect empty: they all have the neutral element as a common element. Anyway the question is too broad. You can't really understand the significance just by looking at definitions. For example if you look at $G/H$ as a $G$-set (i.e. a set equiped with an action of $G$) then $G/H$ is $G$-isomorphic to $G/K$ if and only if $H$ is conjugate to $K$. This is deeply related to the representation theory of groups, e.g. the Burnside ring. – freakish Dec 7 '17 at 19:54
• How distracted I must be to have forgotten that! Alright then must the elements of $K$ and$H$ be the same? How do you calculate how many conjugate subgroups there are in a Group? – Kam Dec 7 '17 at 20:01
• @Kam This is hard. There is no known efficient way to do that, even when $G$ is finite. Read more here: groupprops.subwiki.org/wiki/Finding_conjugate_subgroups – freakish Dec 7 '17 at 20:04

I know that there is an equivalence relation on the conjugacy of subgroups of $G$, so they partition $G$.

No, they don't. The relation is defined over the set of all subgroups. Not over $G$ itself.

Question 1: What significance do Invariant Subgroups have under automorphisms?

You mean characteristic subgroup? Read more here: https://en.wikipedia.org/wiki/Characteristic_subgroup

Question 2: What does it mean for two subgroups to be a conjugate of one another.

Subgroup $H$ is conjugate to subgroup $K$ if and only if there is $g\in G$ such that $gHg^{-1}=K$. By definition $H$ is normal if and only if it has no conjugates except for itself.

Question 3: What is the importance of partitioning $G$ with groups as opposed to cosets?

What partitioning? The equivalence relation "$H$ is conjugate to $K$" is an equivalence relation over the set of all subgroups of $G$. Not on $G$.

Anyway the conjugacy relation is useful. It is used mainly in the representation theory of groups (e.g. the Burnside ring).

• This helps a lot! Thanks again – Kam Dec 7 '17 at 20:09
• Forgive me for persisting, but if "$H$" is conjugate to $K$" is an equivalence relation over the set of all subgroups of $G$, then what is partitioned and how is it partitioned? – Kam Dec 7 '17 at 22:14

The "equiv rel on the conjugacy of subgroups" doesn't introduce a partition on $G$ but on the set of subgroups of $G$.

Normal Subgroups allow their cosets to form a group using the natural multiplication operation.

Orbit is a generalization of conjugacy class; orbit refers to the set of images under a permutation group of any set, conjugacy class is the orbit of a subset of a group under conjugation. The subset could be a single element, a subgroup, or anything else.

Each conjugation permutes the elements of $G$; they are also homomorphisms; so they are isomorphisms of $G\to G$. An isomorphism of $G\to G$ is called an automorphism. There are automorphisms that do not arise this way, The set of all automorphisms forms a group under composition.

The stabilizer is the subset (it is a subgroup) of a permutation group acting on a set, $S$ which maps $S \to S$.

"What is the connection?" To a significant part, this is the goal of the study of group theory.

You have asked a multipart question and many follow up questions. I applaud your inquisitiveness. We could go on for a long time and that might be fun, but.... I have a feeling though that this is very abstract to you.

I suggest that you play around with a specific group for a while to help solidify these ideas. A good group to start with might be $$D_4 \text{ the Dihedral group of order 8}$$ It would be good if you didn't look it up but worked with it yourself until you understand the concepts you have been questioning us about. The elements of the group are $$\{a^{\epsilon}b^i | \epsilon \in \{0,1 \} \text { and } i \in \{0,1,2,3 \}$$ and multiplication can be derived from the following relations: $$a^2 = b^4 = (ab)^2 = e \text { the identity element}$$ The group has 3 subgroups of order 4 and 5 subgroups of order 2. Have fun.

• Since $\mathbb Z/ 12\mathbb Z$ is Abelian (everything commutes) all cojugacy classes are singletons. – Stephen Meskin Dec 7 '17 at 20:06
• "Could you just explain how each normal subgroup is partitioned?" This question is not meaningful to me. – Stephen Meskin Dec 7 '17 at 20:10
• What exactly is partitioned then with the equivalence relation, and how? – Kam Dec 7 '17 at 20:11
• Which E.R. on which set? – Stephen Meskin Dec 7 '17 at 20:15
• "The "equiv rel on the conjugacy of subgroups" doesn't introduce a partition on G but on the subgroups of G." – Kam Dec 7 '17 at 20:16