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Questions tagged [quotient-group]

This tag is for questions relating to "Quotient Group".

A quotient group or factor group is a group obtained by aggregating similar elements of a larger group using an equivalence relation that preserves the group structure.

Definition: If $G$ is a group and $N$ is a normal subgroup of group $G$, then the set $G/N$ of all cosets of $N$ in $G$ is a group with respect to the multiplication of cosets. It is called the quotient group or factor group of $G$ by $N$. The identity element of the quotient group $G/N$ by $N$.

  • Any normal subgroup has a corresponding quotient group, formed from the larger group by eliminating the distinction between elements of the subgroup.
  • In category theory, quotient groups are examples of quotient objects, which are dual to subobjects.
  • The quotient group construction can be viewed as a generalization of modular arithmetic to arbitrary groups. In fact, the quotient group $G/N$ is read "$G$ mod $N.$"
  • It can be verified that the set of self-conjugate elements of $G$ forms an abelian group $Z$ which is called the center of $G$.

References:

https://en.wikipedia.org/wiki/Quotient_group

https://brilliant.org/wiki/quotient-group/

914 questions
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Why do we define quotient groups for normal subgroups only?

Let $G \in \mathbf{Grp}$, $H \leq G$, $G/H := \lbrace gH: g \in G \rbrace$. We can then introduce group operation on $G/H$ as $(xH)*(yH) := (x*y)H$, so that $G/H$ becomes a quotient group when $H$ is a normal subgroup. But why do we only work with…
Aleksei Averchenko
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Does $G\cong G/H$ imply that $H$ is trivial?

Let $G$ be any group such that $$G\cong G/H$$ where $H$ is a normal subgroup of $G$. If $G$ is finite, then $H$ is the trivial subgroup $\{e\}$. Does the result still hold when $G$ is infinite ? In what kind of group could I search for a…
Klaus
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Isomorphic quotient groups

Suppose that $G$ is a finite group, and that $H$ and $K$ are normal subgroups of $G$ with trivial intersection, and suppose that $H$ and $K$ are isomorphic. Is it true that the quotient groups $G/H$ and $G/K$ are isomorphic?
Iota
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Finite group with isomorphic normal subgroups and non-isomorphic quotients?

I know it is possible for a group $G$ to have normal subgroups $H, K$, such that $H\cong K$ but $G/H\not\cong G/K$, but I couldn't think of any examples with $G$ finite. What is an illustrative example?
Zev Chonoles
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Why the term and the concept of quotient group?

The basic concept of Quotient Group is often a confusing thing for me,I mean can any one tell the intuitive concept and the necessity of the Quotient group, I thought that it would be nice to ask as any basic undergraduate can learn the intuition…
IDOK
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4 answers

Give an example of: A group with an element A of order 3, an element B with order 4, where order of AB is less than 12

I'm a mathematics major studying at University as an undergrad. This is a question on the study guide for the upcoming final in Math 344 - Group Theory: "Give an example of a group G with an element a of order 3, an element b of order 4, where…
Jay Ess
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If $H$ is a subgroup of a finite abelian group $G$, then $G$ has a subgroup that is isomorphic to $G/H$.

I know Is every quotient of a finite abelian group $G$ isomorphic to some subgroup of $G$? has two answers. I don't understand how the first answer works and I have doubt about that answer. The second answer uses character theory which I don't…
Collier Gaiser
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14
votes
4 answers

Difference between $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}_n$

In every Modern Algebra book I've read, I've seen that the groups $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}_n$ are isomorphic, but not equal. I understand the difference between "isomorphic" and "equal," but this particular example raises a couple of…
d4rk_1nf1n1ty
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Is $\Bbb R/\Bbb Z$ isomorphic to $\Bbb R/2\Bbb Z$?

I see that both ${\Bbb R}/{\Bbb Z}$ and ${\Bbb R}/{2\Bbb Z}$ are isomorphic to $S^1$. But when I apply the third isomorphism theorem I get ${\Bbb R}/{\Bbb Z}\simeq\frac {{\Bbb R}/{2\Bbb Z}}{{\Bbb Z}/{2\Bbb Z}}$ i.e. ${\Bbb R}/{\Bbb Z}\simeq\frac…
Smooth Alpert Frame
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7 answers

Why is the fact that a quotient group is a group relevant?

I'm studying the basics of quotient groups. I understand that if you build a quotient set from cosets of a group and the subgroup you are using to build them is normal then you end up with a group. I fail to see why the fact that we can define…
Joaquin Brandan
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5 answers

If $G$ is cyclic then $G/H$ is cyclic?

If $G$ is cyclic, then $G/H$ is cyclic? The proof I got goes like this: $G$ is cyclic, so $G=$ for some $g\in G$. So any coset in $G/H$ would be of the form $Hg'=Hg^n$ for some $n$. So $Hg$ is an generator of $G/H$. Thus, $G/H$ is cyclic. I…
3x89g2
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Are Subgroups' Quotients also Quotients' Subgroups?

Let $G$ be a group, $G'\le G$ a subgroup, and $\varphi:G'\to H$ a surjective group homomorphism. Must there exist a surjective group homomorphism $\psi:G\to H'$ such that $H'\ge H$? I know the converse is true: Take $G'=\psi^{-1}(H)$ and let…
Edward H
  • 142
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2 answers

$H$ is a maximal normal subgroup of $G$ if and only if $G/H$ is simple.

I need to prove that $H$ is a maximal normal subgroup of $G$ if and only if $G/H$ is simple. My proof to the $(\Rightarrow)$ direction seems too much trivial: Let us assume there exist $A$ so that $A/H\lhd G/H$. Then by definion, $H$ must be normal…
catch22
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10
votes
2 answers

Is this explanation of normal subgroups and quotient groups correct?

I apologize for the long post, but I'm currently a student finishing up his first semester in group theory. My introduction was pretty definition-heavy so I've found I can internalize concepts (such as quotient groups, normal subgroups, etc.) myself…
Andrew Li
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10
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7 answers

Conditions for cyclic quotient group

Let $G$ be an arbitrary finite group and $H$ a normal subgroup. What are some good conditions on $H$ that make the quotient $G/H$ cyclic? I want to avoid any further restriction on $G$.
Samuel Plath
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