In any finite group, number of elements not equal to their own inverses is even number  In my book they have paired elements with their inverses, being elements and inverses different from each other. How do i see this hint ? Thanks
120710
The problem I am attempting to prove is the following: In any finite group $G$, the number of elements not equal to their own inverse is an even number.  Caveat: I have had very limited experience with proofs. Any comments would be appreciated.  Let $a_{1}, a_{2}, ..., a_{n}$ be elements of $G$. Since $G$ is a group, every element in $G$ must have an inverse in $G$. Let the inverse of each element be represented by $a_{1}^{-1}, a_{2}^{-1}, ..., a_{n}^{-1}$. An element $a_{k}$ in $G$ that is not equal to its own inverse can be written as $a_{k} \neq a_{k}^{-1}$. For every element $a_{k}$, there is another element $a_{k}^{-1}$ that is not equal to its own inverse denoted by $a_{k}^{-1} \neq a_{k}$. So for every element $a_{k}$ in $G$ whose inverse is not equal to their own inverse, there will be 2 elements in $G$ namely $a_{k}$ and $a_{k}^{-1}$. Let the integer $q$ denote the number of elements ranging from $a_{1}, a_{2}, ..., a_{n}$ who are not equal to their own inverse. In addition for every value $q$, there must be another element in $a_{1}^{-1}, a_{2}^{-1}, ..., a_{n}^{-1}$ not equal to their own inverse. So the total number of elements not equal to their own inverse can be denoted by 2$q$, which is an even number.
63864
If a set has $n$ elements, how many transitive relations are there on it?  For example if set $A$ has $2$ elements then how many transitive relations. I know the total number of relations is $16$ but how to find only the transitive relations? Is there a formula for finding this or is it a counting problem?  Also how to find this for any number of elements $n$?
17353
Please note: I posted this question about an hour ago on my account (I just made it as I am new to the site). I revisited my post and saw that I had not given definitions of what I was asking (I thought they were common notations) and it would not let me edit the post for some reason so I will ask here now and hopefully someone can delete the other thread.  How to prove for a finite group $G$ that $|\langle g \rangle|=o(g)$?  Here are the definitions:  Let $G$ be a finite group and $g \in G$  $o(g)$ is the smallest $m \in \Bbb{N}$ such that $g^m=e$ (the identity element of the group). (If such an $m$ exists)  $\langle g \rangle=\{g^m : m \in \mathbb{Z}\}$  I really don't know how to show that this is true I tried to say let $o(g)=m$ and then show that $|\langle g \rangle|$ has exactly $m$ elements but I wasn't even sure it did because it would contain the elements $g^0,...g^{m-1}$ so that would mean it's true but what about $g^{-1}$ and the other elements with negative coefficients so it seems like it would actually be bigger than $m$?  Thanks.
54138
I'm working through Rotman (1994). It's the 2nd chapter where my curiosity lead me to a question that is not from the book. Here's the problem that inspired the question:      2.8. Imbed $S_n$ as a subgroup of $A_{n+2}$, but show, for $n\ge2$, that $S_n$ cannot be imbedded in $A_{n+1}.$    Since the alternate subgroup has the property of even parity, I thought that a little investigation of parity for the second and third symmetric groups was in order. So I constructed a table for each of them, grouped by even and odd number of disjoint products.   I'll forego writing the actual table and simply provide the answers. In symmetric groups with an even idegree, two even permutations come together under composition to produce an even permutation. Two odds also make an even. Two different parities produce odd elements.   When the symmetric group has an odd degree, e.g. $S_3$, two odds make an odd, two evens make an odd, and two different parities make an even element. Exactly the opposite results from the symmetric groups with even index.   Two questions: is this enough to prove the $A_{n+1}$ case? And can this be generalized modulo 2 e.g. $A_{2n+1}$?
123245
Find an infinite group, in which every element g not equal identity (e) has order 2   Does this question mean this:  the group that fail condition (2) which is no inverse and also that group must have the size 2  My answer:  Z*
41334
Suppose we are given a finite group $G$, and suppose we know the order of $G$ as well as the orders of each of its elements.  Does this information alone uniquely determine the group up to isomorphism?  What if we add extra hypotheses about $G$ (solvable, simple, etc)?  If not, can we at least extract some useful information regarding its structure?    For example, if $|G| = 8$ and $G$ contains three elements of order $2$ and four elements of order $4$, then I know $G \cong \mathbb{Z}_4\times\mathbb{Z}_2$.
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In our 6th grade math class we got introduced to the concept of integers. With all the talk about positive and negative, it got me wondering. Is the amount of elements in $\mathbb{Z^+}$ less than the amount of elements in $\mathbb{Z}$?  Here is how I thought of it. If we have $\mathbb{Z^+}$ and add one element to the "back" of it ($\mathbb{Z}^{\geq-1}$) there are certainly more elements in that new set, so there must be more elements in $\mathbb{Z}$ than in $\mathbb{Z}^+$ but on the other side if we try to express the amount of elements in them "numerically" (In a loose sense of the word) they both have $\infty$ elements.    So what is the right answer? Is there even one?  (Please note I am a bit of a layman when it comes to mathematics)  Edit: About the possible duplicate, I am not looking for a bijection between the two sets, I am looking for if they even have the same number of elements and why a bijection shows that they do/don't
102223
I was reading Nielsen and Chuang's "Quantum Computation and Quantum Information" and in the appendices was a group theory refresher. In there, I found this question:      Exercise A2.1 Prove that for any element $g$ of a finite group, there always exists a positive integer $r$ such that $g^r=e$. That is, every element of such a group has an order.    My first thought was to look at small groups and try an inductive argument. So, for the symmetric groups of small order e.g. $S_1, S_2, S_3$ the integer $r$ is less than or equal to the order of the group. I know this because the groups are small enough to calculate without using a general proof.   For example, in $S_3$ there is an element that rearranges the identity $\langle ABC \rangle$ element by shifting one character to the left e.g. $s_1 = \langle BCA \rangle$. Multiplying this element by itself produces the terms $s_1^2 = \langle CAB \rangle$; and $s_1^3 = \langle ABC \rangle$ which is the identity element, so this element is equal to the order of the group, which is three.   I have no idea if this relation holds for $S_4$ which means I am stuck well before I get to the general case.   There's a second question I'd like to ask related to the first. Is the order or period of any given element always less than or equal to the order of the group it belongs to?
120865
Show that a group with $|G| = 33$ contains an element of order $3$.   Using Lagrange's theorem, the possible orders are $1$,$3$,$11$, and $33$ (from the factors of $33$).  If there is a subgroup containing an element with an order of $33$, then we are done.   There is only one identity element with an order of $1$.   I am currently left with elements of orders $3$ or $11$. What is the best way to show that any element of order $11$ is a contradiction?
79589
Is there an efficient way to enumerate the unique subgroups of the symmetric group? Naïvely, for the symmetric group $S_n$ of order $\left | S_n \right | = n!$, there are $2^{n!}$ subsets of the group members that could potentially form a subgroup. In addition, many of these subgroups are going to be isomorphic to each other. I feel that the question has an easy answer in terms of the conjugacy classes, but I don't see how. If the answer generalizes to all finite groups, please elaborate!
114711
Let $G = \{e, a_1, a_2, \dots, a_n\}$ be a finite abelian group and let $S$ be the set of all the elements of $G$ which are not equal to their own inverse. The set $S$ can be divided up into pairs so that each element is paired off with its own inverse. Prove that $(a_1 a_2 \dots a_n)^2 = e$.   Every $x$ in $G$, outside $S$ abides by $x^2 = e.$   Let $a_i$ be in $S$. Since $G$ is abelian, we have  $a_ia_i^{-1} = a_i^{-1}a_i$  $e = a_i^{-1}a_i$  Since $a_i^{-1} = a_i$, $a_i^2 = e.$  Thus, $\dots a_i^2a_j^2\ldots a_n^2 = e.$  Does that make sense?  edit:  We need to prove $a_1a_2...a_na_1a_2...a_n = e$, right? Let $a_i$ be in $G \setminus S$. Then $(a_i)^2$ is $e$, is it? If we rearrange the elements on the left side of the equation above according to what belongs to $S$ and what belongs to $G \setminus S$ and take the product of all the $(a_i)^2$, then every element belonging to $G \setminus S$ will  be cancelled in the right side of $a_1a_2...a_na_1a_2...a_n = e$. So, we only need to consider the elements in $S$. Suppose $a_j \in S$. Let $a_k$ be the inverse of $a_j$. We need to prove $(a_j)^2 \cdot (a_k)^2 = a_j \cdot a_j \cdot a_k \cdot a_k = e$. Since $S$ is abelian , $a_j \cdot a_j \cdot a_k \cdot a_k$ can be rearranged as $a_j \cdot a_k \cdot a_j \cdot a_k$ which is $e$.
119250
Given a set of whole numbers of size $n$, how many unique combinations of these whole numbers are there?   For example, for the set $\{1,2,3\}$ the unique combinations are   $\{1,2,3\}, \{1,2\}, \{1\}, \{2\}, \{3\}, \{2,3\}, \{1,3\}$   The number of combinations for $\{1,2,3\}$ is $7$.  Does this generalize to the number of unique combinations in a set is $2^n-1$ where $n$ is the number of elements in the set?
74322
If a cyclic group has an element of infinite order, how many elements of finite order does it have?  I know that the order of the entire group must be infinite, for an element of the group must have an order less than the group order. My first thought was that there are no elements with finite order in this group, however now I'm believing that there are infinitely many elements of finite order, since the group should have infinitely many elements. Can someone explain the solution?
82188
I recently took a course on group theory, which mentioned that the following proposition is equivalent to the continuum hypothesis: "The infinite symmetric group (i.e. the group of permutations on the set $\mathbb{N}$) has exactly 4 normal subgroups." Does anyone have any references or explanation for this?
76923
Suppose that a set has an odd number of elements. Explain why half of the subsets will have an odd number of elements .  Now assuming set A is the set with an odd number of integer elements{1,2,3,4,5}  Subset b ={1,2}  subset c={3}  Subset d={4,5}  Now there are two subsets with even number of elements and one subset with an odd number of elements.It seems to contradict the earlier theorem that i have to prove.I think that i am missing something.
52208
If $G$ has at least $7$ elements of order $7$, then prove that $G$ has at least $12$ elements of order $7$.   I know that an element having order $7$ means that for any $a$ in $G$, $a^7=e$ where $e$ is the identity element. The inverses also have same order, so I thought maybe of the $7$ elements, $6$ had their inverse as well. However, then this would just mean there had to be $8$ elements, not $12$ for the $7$th element to have it's inverse.
91092
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 must divide the order of $G$, but it is odd), and so every element except $e$ has a unique inverse which is different from it. Hence both the element and its inverse participate in the product and cancel each other.  My problem is simple - I need to solve this without Lagrange's theorem. So either there's a smart way to prove the nonexistance of an element of order 2 in an odd abelian group, or I'm missing something even more basic...