A group is an algebraic structure consisting of a set of elements together with an operation that satisfies four conditions: closure, associativity, identity and invertibility. Group theory studies groups.

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Does the intersection of two finite index subgroups have finite index?

Let $(G,*)$ be a group and $H,K$ be two subgroups of $G$ of finite index (the number of left cosets of $H$ and $K$ in $G$). Is the set $H\cap K$ also a subgroup of finite index? I feel like need that $...
1
vote
2answers
425 views

Showing that $G$ is a group under an alternative operation.

Let $G$ be a group and let $c$ be a fixed elements of $G$. Now, I'm going to define a new operation "*" on $G$ by $a*b=ac^{-1}b$ How do I prove that the set $G$ is a group under *. Thanks for ...
84
votes
2answers
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More than 99% of groups of order less than 2000 are of order 1024?

In Algebra: Chapter 0, the author made a remark (footnote on page 82), saying that more than 99% of groups of order less than 2000 are of order 1024. Is this for real? How can one deduce this result? ...
16
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2answers
886 views

Has this “generalized semidirect product” been studied?

If $G$ is a finite group with subgroups $H$ and $K$ such that $HK = G$ and $H\cap K = \{1\}$ we get that every element of $G$ can be written uniquely as $hk$ with $h\in H$ and $k\in K$. This then ...
12
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3answers
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A normal subgroup intersects the center of the $p$-group nontrivially

If $G$ is a finite $p$-group with a nontrivial normal subgroup $H$, then the intersection of $H$ and the center of $G$ is not trivial.
26
votes
7answers
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Product of all elements in an odd finite abelian group is 1

This should be an easy exercise: Given a finite odd abelian group $G$, prove that $\prod_{g\in G}g=e$. Indeed, using Lagrange's theorem this is trivial: There is no element of order 2 (since the order ...
21
votes
1answer
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Is it true that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as abelian groups?

I think the answer is yes. Sketch of the proof Consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$. Let $\{e_\lambda:\lambda\in\Lambda\}\subset\mathbb{R}$ be its Hamel basis. Then $\{(e_{\...
17
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2answers
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Are normal subgroups transitive?

Suppose $G$ is a group and $K\lhd H\lhd G$ are normal subgroups of $G$. Is $K$ a normal subgroup of $G$, i.e. $K\lhd G$? If not, what extra conditions on $G$ or $H$ make this possible? Applying the ...
9
votes
4answers
405 views

If $[G:H]=n$, is it true that $x^n\in H$ for all $x\in G$?

Let $G$ be a group and $H$ a subgroup with $[G:H]=n$. Is it true that $x^n\in H$ for all $x\in G$? Remarks. The answer is positive whenever $H$ is normal, e.g., for $n=2$. In general, by using the ...
12
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3answers
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Is a semigroup $G$ with left identity and right inverses a group?

Hungerford's Algebra poses the question: Is it true that a semigroup $G$ that has a left identity element and in which every element has a right inverse is a group? Now, If both the identity and the ...
3
votes
3answers
843 views

Cancellation of Direct Products

Given a finite group $G$ and its subgroups $H,K$ such that $$G \times H \cong G \times K$$ does it imply that $H=K$. Clearly, one can see that this doesn't work out for all subgroups. Is there any ...
7
votes
2answers
1k views

Product of elements of a finite abelian group

Suppose $G=\{a_1,...,a_n\}$ is a finite abelian group, and let $x=a_1a_2\dotsm a_n$. Prove that if there is more than one element of order $2$ then $x=e$. What I've done so far: (#1 is just for ...
6
votes
3answers
2k views

Finite Index of Subgroup of Subgroup

Prove the following: If $H$ is a subgroup of finite index in a group $G$, and $K$ is a subgroup of $G$ containing $H$, then $K$ is of finite index in $G$ and $[G:H] = [G:K][K:H]$. So this is ...
13
votes
1answer
595 views

If a group satisfies $x^3=1$ for all $x$, is it necessarily abelian?

I know that any group satisfying $x^2=1$ for all $x$ is abelian. Is the same true if $x^3=1$? I don't think it is, but I can't find a basic counterexample.
38
votes
4answers
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Center-commutator duality

I'm reading this article by Keith Conrad, on subgroup series. I'm having trouble with a statement he does at page 6: Any subgroup of $G$ which contains $[G,G]$ is normal in $G$. He says this as ...
62
votes
4answers
8k views

The direct sum $\oplus$ versus the cartesian product $\times$

In the case of abelian groups, I have been treating these two set operations as more or less indistinguishable. In early mathematics courses, one normally defines $A^n := A\times A\times\ldots\times A$...
26
votes
2answers
2k views

Yoneda-Lemma as generalization of Cayley`s theorem?

I came across the statement, that Yoneda-lemma is a generalization of Cayley`s theorem which states, that every group is isomorphic to a group of permutations. How exactly does generalizes Yoneda-...
27
votes
14answers
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19
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4answers
3k views

Derived subgroup where not every element is a commutator

Let $G$ be a group and let $G'$ be the derived subgroup, defined as the subgroup generated by the commutators of $G$. Is there an example of a finite group $G$ where not every element of $G'$ is a ...
14
votes
1answer
639 views

Does $gHg^{-1}\subseteq H$ imply $gHg^{-1}= H$? [duplicate]

Let $G$ be a group, $H<G$ a subgroup and $g$ an element of $G$. Let $\lambda_g$ denote the inner automorphism which maps $x$ to $gxg^{-1}$. I wonder if $H$ can be mapped to a proper subgroup of ...
9
votes
2answers
4k views

How to prove that if $G$ is a group with a subgroup $H$ of index $n$, then $G$ has a normal subgroup $K\subset H$ whose index in $G$ divides $n!$

I would be very thankful if someone could give me a hint with proving next statement: If $G$ is a group with a subgroup $H$ of finite index $n$, then $G$ has a normal subgroup $K$ contained in $H$ ...
16
votes
2answers
780 views

Number of finite simple groups of given order is at most $2$ - is a classification-free proof possible?

This Wikipedia article states that the isomorphism type of a finite simple group is determined by its order, except that: $L_4(2)$ and $L_3(4)$ both have order $20160$ $O_{2n+1}(q)$ and $S_{2n}(q)$ ...
13
votes
1answer
3k views

Centralizer of a given element in $S_n$?

It is known that any two disjoint cycles in $S_n$ commutes. Therefore, any $\pi\in S_n$ which is disjoint with $\sigma$ is in the centralizer of $\sigma$: $C_{S_n}(\sigma)$. Also $$ \sigma^i\pi\in C_{...
12
votes
4answers
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A kind of converse of Lagrange's Theorem

Let $G$ a finite group. If $a\in G$, a consequence of the Lagrange's Theorem is that the order of $a$ divides the order of $G$. Let $p=|G|$. If $p$ is prime, it is well known that $G$ is cyclic, and ...
10
votes
3answers
1k views

Classifying the irreducible representations of $\mathbb{Z}/p\mathbb{Z}\rtimes \mathbb{Z}/n \mathbb{Z}$

I need help with the following problem: Suppose $n$ is some positive integer, and $n|p-1$. Classify all irreducible representations of $$\mathbb{Z}/p\mathbb{Z}\rtimes \mathbb{Z}/n \mathbb{Z}.$$ ...
8
votes
3answers
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Show group of order $4n + 2$ has a subgroup of index 2.

Let $n$ be a positive integer. Show that any group of order $4n + 2$ has a subgroup of index 2. (Hint: Use left regular representation and Cauchy's Theorem to get an odd permutation.) I can easily ...
7
votes
2answers
422 views

Does $G\times K\cong H\times K$ imply $G\cong H$?

Let $G\times K$ be a finite group. Suppoose $G\times K\cong H\times K$. Is this sufficient to imply $G\cong H$?
12
votes
4answers
771 views

Elementary isomorphism between $\operatorname{PSL}(2,5)$ and $A_5$

At this Wikipedia page it is claimed that to construct an isomorphism between $\operatorname{PSL}(2,5)$ and $A_5$, "one needs to consider" $\operatorname{PSL}(2,5)$ as a Galois group of a Galois cover ...
6
votes
2answers
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Nonabelian semidirect products of order $pq$?

I just constructed the semidirect product in Lang, and I'm trying to tie some facts together. From Ash's Algebra, I know that if $p\lt q$ are distinct primes, if $q\not\equiv 1\pmod{p}$, then any ...
12
votes
1answer
497 views

When a semigroup can be embedded into a group

Under what assumptions can a semigroup $(S,*)$ be embedded into a group?
8
votes
2answers
255 views

Which groups are derived subgroups?

Let $G$ be a group. When is there a group $H$ such that $G$ is isomorphic to its derived subgroup $H'$? I only know that there is not always such a $H$; for instance, no group has its derived ...
7
votes
2answers
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An element of a group has the same order as its inverse

If $a$ is a group element, prove that $a$ and $a^{-1}$ have the same order. I tried doing this by contradiction. Assume $|a|\neq|a^{-1}|$ Let $a^n=e$ for some $n\in \mathbb{Z}$ and $(a^{-1})^m=e$...
7
votes
1answer
599 views

What is $\operatorname{Aut}(\mathbb{R},+)$?

I was solving some exercises about automorphisms. I was able to show that $\operatorname{Aut}(\mathbb{Q},+)$ is isomorphic to $\mathbb{Q}^{\times}$. The isomorphism is given by $\Psi(f)=f(1)$, but ...
6
votes
1answer
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Normal subgroups of dihedral groups

In relation to my previous question, I am curious about what exactly are the normal subgroups of a dihedral group $D_n$ of order $2n$. It is easy to see that cyclic subgroups of $D_n$ is normal. But ...
8
votes
6answers
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In what sense of “structure” do group homomorphisms “preserve structure”?

It is commonly said that group homomorphisms "preserve the structure of the group", e.g., from Wikipedia: The purpose of defining a group homomorphism as it is, is to create functions that ...
9
votes
3answers
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Prove that the center of a group is a normal subgroup

Let $G$ be a group. We define $H$ where $H$ is the center of/centralizer of $G$: $$H=\{h\in G| \forall g\in G: hg=gh\}.$$ Prove that $H$ is a (normal) subgroup of $G$.
1
vote
3answers
237 views

An example of a residually finite group which is not Hopf

trying to think of any residually finite group which is not Hopf. Any help?
63
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4answers
2k views

How is a group made up of simple groups?

I've read more than once the analogy between simple groups and prime numbers, stating that any group is built up from simple groups, like any number is built from prime numbers. I've recently started ...
32
votes
8answers
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Is symmetric group on natural numbers countable?

I guess it is too difficult a question to ask about the cardinality of $S_{\mathbb{N}}$ so I would like to ask whether it is countable or not. I tried to prove it is uncountable somewhat mimicking ...
15
votes
5answers
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How can you show there are only 2 nonabelian groups of order 8?

It's often said that there are only two nonabelian groups of order 8 up to isomorphism, one is the quaternion group, the other given by the relations $a^4=1$, $b^2=1$ and $bab^{-1}=a^3$. I've never ...
15
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2answers
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$A_n$ is the only subgroup of $S_n$ of index $2$.

How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$? Why isn't there other possibility? Thanks :)
14
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Theorems with the greatest impact on group theory as a whole

In his Contemporary Abstract Algebra text, Gallian asserts that Sylow's Theorem(s) and Lagrange's Theorem are the two most important results in finite group theory. He also provides this quote by G.A....
18
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1answer
522 views

A special subgroup of groups of order $n$

Let $G$ be a group with $|G| = n$ and let $ \emptyset \ne S \subseteq G$. I want to show that $S^n$ is a SUBGROUP of $G$ where by $S^n$ I mean the set $\lbrace s_1\cdots s_n \; | \; s_i \in S\rbrace$.
13
votes
3answers
534 views

If $|\lbrace g \in G: \pi (g)=g^{-1} \rbrace|>\frac{3|G|}{4}$, then $G$ is an abelian group.

Assume that $\pi$ is an automorphism of a finite group $G$. Let $S$ denote the set $\lbrace g \in G: \pi (g)=g^{-1} \rbrace$. Show that if $|S|>\frac{3|G|}{4}$, then $G$ is an abelian group. ...
10
votes
5answers
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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 ...
10
votes
4answers
4k views

Automorphism group of the quaternion group

Let $Q_8$ be the quaternion group. How do we determine the automorphism group $Aut(Q_8)$ of $Q_8$ algebraically? I searched for this problem on internet. I found some geometric proofs that $Aut(Q_8)$ ...
26
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2answers
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Finite number of subgroups $\Rightarrow$ finite group

I'm trying to prove that any group $G$ of infinite order has an infinite number of subgroups. I think that if the group has an element of infinite order, then it's easy because I can take the groups ...
17
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2answers
3k views

Is every quotient of a finite abelian group $G$ isomorphic to some subgroup of $G$?

I'm having difficulty with exercise 1.43 of Lang's Algebra. The question states Let $H$ be a subgroup of a finite abelian group $G$. Show that $G$ has a subgroup that is isomorphic to $G/H$. ...
19
votes
3answers
2k views

Subgroup of $\mathbb{Q}$ with finite index

Consider the group $\mathbb{Q}$ under addition of rational numbers. If $H$ is a subgroup of $\mathbb{Q}$ with finite index, then $H = \mathbb{Q}$. I just saw this on our exam earlier and was stumped ...
12
votes
5answers
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Why is the set of commutators not a subgroup?

I was surprised to see that one talks about the subgroup generated by the commutators, because I thought the commutators would form a subgroup. Some research told me that it's because commutators are ...