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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498 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 ...
15
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2answers
782 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 ...
7
votes
4answers
2k views

Order of a product of subgroups

Let $H$, $K$ be subgroups of $G$. Prove that $o(HK) = \frac{o(H)o(K)}{o(H \cap K)}$. I need this theorem to prove something.
23
votes
2answers
2k views

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 ...
23
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7answers
3k views

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 ...
13
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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 $G/H$. ...
7
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3answers
1k views

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 ...
8
votes
1answer
235 views

Isomorphism between $I_G/I_G^2$ and $G/G'$

Ok, this has been bugging me for a while, and I'm sure there's something obvious I'm missing. The references I've looked at for this result in an effort to resolve the issue didn't address it. $G$ is ...
7
votes
2answers
3k views

Show that every group of prime order is cyclic

Show that every group of prime order is cyclic. I was given this problem for homework and I am not sure where to start. I know a solution using Lagrange's theorem, but we have not proven ...
6
votes
3answers
336 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 $G\cong H$?
2
votes
1answer
311 views

Another point of view that $\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}$ is cyclic.

I was thinking that the product of groups $\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/2\mathbb{Z}$ is not cyclic, but $\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/p\mathbb{Z}$ is cyclic if p is an odd prime. ...
7
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3answers
8k views

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 $G$.
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3answers
187 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?
1
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4answers
2k views

Need to prove that (S,*) defined by the binary operation a*b = a+b+ab is an abelian group on S = R \ {1}

So basically this proof centers around proving that (S,*) is a group, as it's quite easy to see that it's abelian as both addition and multiplication are commutative. My issue is finding an identity ...
59
votes
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 ...
18
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 ...
20
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14answers
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17
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7answers
6k views

How to show every subgroup of a cyclic group is cyclic?

I'm teaching a group theory course now, and I wanted to give my students a proof that every subgroup of a cyclic group is cyclic. The easiest way I could think to do this is to say that any cyclic ...
15
votes
7answers
1k views

Why $PSL_3(\mathbb F_2)\cong PSL_2(\mathbb F_7)$?

Why are groups $PSL_3(\mathbb{F}_2)$ and $PSL_2(\mathbb{F}_7)$ isomorphic? Update. There is a group-theoretic proof (see answer). But is there any geometric proof? Or some proof using octonions, ...
14
votes
2answers
1k views

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 ...
19
votes
2answers
821 views

Is every group the automorphism group of a group?

Suppose $G$ is a group. Does there always exist a group $H$, such that $\operatorname{Aut}(H)=G$, i. e. such that $G$ is the automorphism group of $H$? EDIT: It has been pointed out that the answer ...
12
votes
4answers
1k views

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 ...
9
votes
4answers
3k 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 $Aut(Q_8)$ ...
8
votes
2answers
1k views

A Nontrivial Subgroup of a Solvable Group

Question: Let $G$ be a solvable group, and let $H$ be a nontrivial normal subgroup of $G$. Prove that there exists a nontrivial subgroup $A$ of $H$ that is Abelian and normal in $G$. [ref: this is ...
11
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 ...
9
votes
4answers
375 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 ...
2
votes
2answers
302 views

Isomorphic Group with $G=(\mathbb Z_{2^\infty}\oplus \frac{\mathbb Q}{\mathbb Z}\oplus \mathbb Q)\otimes_{\mathbb Z}\mathbb Q $

Let $$G=\left(\mathbb Z_{2^\infty}\oplus\mathbb Q/\mathbb Z\oplus \mathbb Q\right)\otimes_{\mathbb Z}\mathbb Q $$ Now $G$ isomorphic with which case: $0$ ? , or $\mathbb Q \, $ ? , or $\mathbb ...
19
votes
1answer
875 views

Are $(\mathbb{R},+)$ and $(\mathbb{C},+)$ isomorphic as additive groups?

Are $(\mathbb{R},+)$ and $(\mathbb{C},+)$ isomorphic as additive groups? I know that there is a bijection between $\mathbb{R}$ and $\mathbb{C}$, and this question asks whether they are isomorphic as ...
12
votes
3answers
305 views

How to prove that a group with some properties is abelian?

Let $(G,.)$ be a group and $m,n\in\mathbb Z$ such that $\gcd(m,n)=1 $ and $$ \forall a,b \in G:a^mb^m=b^ma^m$$ $$\forall a,b \in G:a^nb^n=b^na^n.$$ Then how prove $G$ is an abelian group ? Thanks ...
10
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2answers
302 views

Known bounds for the number of groups of a given order.

The number of nonisomorphic groups of order $n$ is usually called $\nu(n)$. I found a very good survey about the values. $\nu(n)$ is completely known absolutely up to $n=2047$, and for many other ...
5
votes
2answers
1k views

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
2answers
2k views

Group of positive rationals under multiplication not isomorphic to group of rationals

A question that may sound very trivial, apologies beforehand. I am wondering why $( \mathbb{Q}_{>0} , \times )$ is not isomorphic to $( \mathbb{Q} , + )$. I can see for the case when $( \mathbb{Q} ...
10
votes
1answer
357 views

When a semigroup can be embedded into a group

Under what assumptions can a semigroup $(S,*)$ be embedded into a group?
7
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1answer
787 views

Infinite group with only two conjugacy classes

Can you show me a reasonably simple (using only elementary group-theoretic tools) example of infinite group with just 2 conjugacy classes ?
6
votes
2answers
231 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
6answers
606 views

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 ...
3
votes
3answers
835 views

Image of subgroup and Kernel of homomorphism form subgroups

Is my proof ok? Let $f:G\to G^{\prime}$ be a group homomorphism and let $H\lt G$. $Im(H) = \{f(x):x\in H\}$. To show that $Im(H)$ is a group, it suffices to show that $f(x)f(y)^{-1}\in Im(H)$. ...
3
votes
3answers
297 views

Existence of subgroup of order six in $A_4$

Show that the alternating group $A_4$ of all even permutations of $S_4$ does not contain a subgroup of order $6$. For me am thinking to write all elements of $A_4$ and trying to find every ...
2
votes
3answers
304 views

Relationship between $\operatorname{ord}(ab), \operatorname{ord}(a)$, and $\operatorname{ord}(b)$ [duplicate]

Another homework problem from my Group Theory class. Let $a,b$ be elements of a group, $G$. Let $\operatorname{ord}(a)=m$ and $\operatorname{ord}(b)=n$. Let $a$ and $b$ commute. Prove: If $m$ and ...
6
votes
2answers
505 views

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 ...
4
votes
3answers
156 views

Needing help picturing the group idea.

I have a question; X and Y are subgroups of a group G. If $|X|, |Y| < \infty$, show that $|XY| = \frac{|X||Y|}{|X \cap Y|}$ but I can't really picture what it is talking about to even get ...
4
votes
2answers
552 views

Coloring dodecahedron

I found some months ago that there are the Polya's enumeration theorem to compute number of colorings of dodecahedron. I got interested to find how to show by using only Burnside's lemma that there ...
2
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2answers
253 views

When does the isomorphism $G\simeq ker(\phi)\times im(\phi)$? hold?

Suppose you have a group isomorphism given by the first isomorphism theorem: $G/ker(\phi) \simeq im(\phi)$ What can we say about the group $ker(\phi)\times im(\phi)$? In particular, when does the ...
52
votes
3answers
5k 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 ...
10
votes
1answer
406 views

Involutions and Abelian Groups

Suppose that $ G $ is a finite group where at least three-fourths of the elements are involutions, i.e., $$ |I(G)| \geq \frac{3}{4} |G|. $$ (Here, $ I(G) $ denotes the set of all involutions of $ G $, ...
26
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4answers
4k views

Finite Groups with exactly $n$ conjugacy classes $(n=2,3,…)$

I am looking to classify (up to isomorphism) those finite groups $G$ with exactly 2 conjugacy classes. If $G$ is abelian, then each element forms its own conjugacy class, so only the cyclic group of ...
22
votes
1answer
715 views

Recovering a finite group's structure from the order of its elements.

Suppose you know the following two things about a group $G$ with $n$ elements: the order of each of the $n$ elements in $G$; $G$ is uniquely determined by the orders in (1). Question: How ...
23
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1answer
580 views

“Efficient version” of Cayley's Theorem in Group Theory

I'm considering finite groups only. Cayley's theorem says the a group $G$ is isomorphic to a subgroup of $S_{|G|}$. I think it's interesting to ask for smaller values of $n$ for which $G$ is a ...
17
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1answer
486 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 $\lbrace s_1\cdots s_n \; | \; s_i \in S\rbrace$. ...
15
votes
2answers
713 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 $S_{2n}(q)$ ...