The study of symmetry: groups, subgroups, homomorphisms, group actions.

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3
votes
3answers
405 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 $f(x)f(y)^{-1}\in Im(H)$. ...
152
votes
21answers
14k views

Nice examples of groups which are not obviously groups

I am searching for some groups, where it is not so obvious that they are groups. In the lectures script there are only examples like $\mathbb{Z}$ under addition and other things like that. I ...
53
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 ...
9
votes
1answer
342 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 $, ...
22
votes
1answer
638 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 ...
14
votes
2answers
622 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: L4(2) and L3(4) both have order 20160 O2n+1(q) and S2n(q) have the same ...
17
votes
7answers
5k 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 ...
13
votes
2answers
623 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 ...
12
votes
1answer
188 views

Problem on abelian group

Let $G$ be an abelian group, and $\Phi:G\to \mathbb{R}$ is a function with the following property: $$\forall a,b\in G,~~ |\Phi(a+b)-\Phi(a)-\Phi(b)|<c$$ The problem asks to prove the existence of ...
12
votes
4answers
2k views

Groups of order $pq$ without using Sylow theorems

If $|G| = pq$, $p,q$ primes, $p \gt q, q \nmid p-1 $, then how do I prove $G$ is cyclic without using Sylow's theorems?
8
votes
4answers
336 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 ...
15
votes
4answers
1k views

Alternative proof that the parity of permutation is well defined?

I learned the following theorem about the properties of permutation from Gallian's Contemporary Abstract Algebra. When I tried to reconstruct the proof myself, I found that it suffices to prove the ...
6
votes
2answers
807 views

If $[G:H]$ and $[G:K]$ are relatively prime, then $G=HK$

I'm struggling to proof that if $H$ and $K$ are subgroups of a finite index of a group $G$ such that [G:H] and [G:K] are relatively prime, then $G=HK$. I don't know why I can't answer it, because this ...
2
votes
2answers
259 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 ...
8
votes
5answers
1k views

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 ...
15
votes
3answers
3k views

Group where every element is order 2

Let $G$ be a group where every non-identity element has order 2. If |G| is finite then $G$ is isomorphic to the direct product $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \ldots \times ...
7
votes
3answers
918 views

Three finite groups with the same numbers of elements of each order

There exist pairs of finite groups $G$ and $H$ such that $G$ and $H$ are not isomorphic, yet they have the same number of elements of each order. For example, if $p$ is an odd prime, then the group ...
6
votes
1answer
618 views

Order of cyclic groups and the Euler phi function

According to Wikipedia, a cyclic number (in group theory) is one which is coprime to its Euler phi function and is the necessary and sufficient condition for any group of that order to be cyclic. Why ...
5
votes
1answer
125 views

If $[G' : G'']\leq p^2$, then $G'$ is abelian.

Problem : Let $G$ be a p-group and $G'$ denote the commutator subgroup of $G$. If $[G' : G'']\leq p^2$, then $G'$ is abelian. It is easy to prove it for the case of $[G' : G'']=1$ since G is ...
2
votes
2answers
499 views

Let $G$ be a finite group and $H\triangleleft G$ a normal subgroup. Prove that $|G/H| =|G|$ if, and only if, $H = \{e\}$.

The group $G$ is a finite group, a group with finite number of elements, and $H\triangleleft G $a normal subgroup. How can we prove that the index $|G/H|=|G|$ iff $H=\{e\}$, the identity element?
3
votes
3answers
728 views

If $a$ and $b$ commute and $\text{gcd}\left(\text{ord}(a),\text{ord}(b)\right)=1$, then $\text{ord}(ab)=\text{ord}(a)\text{ord}(b)$.

Prove if $\operatorname{ord}(a)=m$, $\operatorname{ord}(b)=n$, and $\operatorname{gcd}(m,n)=1$, then $\operatorname{ord}(ab)=mn$. I was reading this and was thinking how this proof would look ...
1
vote
3answers
93 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
vote
1answer
219 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. ...
46
votes
6answers
2k views

What kind of “symmetry” is the symmetric group about?

There are two concepts which are very similar literally in abstract algebra: symmetric group and symmetry group. By definition, the symmetric group on a set is the group consisting of all bijections ...
41
votes
3answers
3k 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 ...
11
votes
2answers
2k views

Is there a systematic way of finding the conjugacy class and/or centralizer of an element?

Is there a systematic way of finding the conjugacy class and centralizer of an element? Could the task be simplified if we are working with "special groups" such as $S_n$ or $A_n$? Are there any ...
14
votes
2answers
308 views

Which finite groups are the group of units of some ring?

Motivated by this question: Which finite groups are the group of units of some ring? Which finite groups are the group of units of some finite ring? Which finite abelian groups are the group of ...
14
votes
4answers
2k 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 ...
19
votes
1answer
315 views

Finite Groups with a subgroup of every possible index

Suppose $G$ is a finite group, with $|G|=n$. Suppose also that for every positive integer $m\mid n$, $G$ has a subgroup of index $m$. Are there any general statements (structural or otherwise) I can ...
13
votes
4answers
1k views

Simple Group Theory Question Regarding Sylow Theorems

It seems that often in using counting arguments to show that a group of a given order cannot be simple, it is shown that the group must have at least $n_p(p^n-1)$ elements, where $n_p$ is the number ...
12
votes
4answers
4k views

Why are two permutations conjugate iff they have the same cycle structure?

I have heard that two permutations are conjugate if they have the same cyclic structure. Is there an intuitive way to understand why this is?
8
votes
2answers
505 views

What is the center of a semidirect product?

Let $G_1$ and $G_2$ be groups. Let $\varphi:G_2\rightarrow \operatorname{Aut}(G_1) $ be a group homomorphism defining the semidirect product $G_1 \rtimes G_2$. Determine the center ...
7
votes
4answers
2k 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 $Aut(Q_8)$ ...
13
votes
4answers
758 views

Coloring the faces of a hypercube

I will restate the 3-D version of the problem. In how many ways can you color a regular cube with 2 colors up to a rotational isometry. The answer is of course a special case of Burnsides Lemma which ...
12
votes
2answers
252 views

Torsion on $\pi_1(X)$, $X$ connected and open in $\mathbb{R}^n$

Can the fundamental group of an open connected subset $X$ of $\mathbb{R}^n$ have a torsion element?
7
votes
2answers
822 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 ...
5
votes
3answers
263 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 $G\cong H$?
10
votes
5answers
328 views

$\varphi$ in $\operatorname{Hom}{(S^1, S^1)}$ are of the form $z^n$

I'd like to see a proof why $\varphi \in \operatorname{Hom}{(S^1, S^1)}$ looks like $z^n$ for an integer $n$. At first I thought I could argue that if I have a homomorphism that maps $e^{ix}$ to some ...
10
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} ...
7
votes
1answer
628 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 ?
7
votes
2answers
338 views

Is there a free subgroup of rank 3 in $SO_3$?

There are known free subgroups of rank 2 in the set of rotations about the origin in $\mathbb{R}^3$, $SO_3$. For instance, the rotations by angle $\arccos \frac {1}{3}$ about the $z$- and $x$-axis ...
6
votes
2answers
2k 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 $H$ ...
6
votes
5answers
4k views

Prove that $(n-1)! \equiv -1 \pmod{n}$ iff $n$ is prime

How can I show that $(n-1)!$ is congruent to $-1 \pmod{n}$ if and only if $n$ is prime? Thanks.
5
votes
3answers
2k views

Conjugacy classes of the nonabelian group of order 21

How would you go about finding the conjugacy classes of the nonabelian group of order 21, $G:=\left\langle x,y | x^7=e=y^3, y^{-1}xy=x^2\right\rangle$?
21
votes
1answer
971 views

Rubik's Cube Not a Group?

I read online that although the 3x3x3 is a great example of a mathematical group, larger cubes aren't groups at all. How can that be true? There is obviously an identity and it is closed, so ...
9
votes
1answer
436 views

The set of all $x$ such that $xHx^{-1}\subseteq H$ is a subgroup, when $H\leq G$

I found this problem in a textbook of abstract algebra: Let $H$ be a subgroup of $G$. Prove that $$\{x\in G:xax^{-1}\in H\text{ for every }a\in H\}$$ is a subgroup of $G$. It's easy to prove ...
8
votes
1answer
621 views

Finite abelian groups - direct sum of cyclic subgroup

Let $G$ be a finite abelian $p$-group. It is quite elementary to see that if $g \in G$ is an element of maximal order (and thus its span is a cyclic subgroup of $G$ of maximal order) then $G$ can be ...
7
votes
1answer
241 views

When a semigroup can be embedded into a group

Under what assumptions can a semigroup $(S,*)$ be embedded into a group?
5
votes
1answer
354 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 ...
5
votes
1answer
2k views

No group of order 36 is simple

Fraleigh(7ed) Example37.14 No group of order 36 is simple. Such a group $G$ has either $1$ or $4$ subgroups of order $9$. If there is only one such subgroup, it is normal in $G$. If there are four ...