Find an infinite group in which every element $g$ not equal to $e$ has order $2$. In this case $e$ is the identity element.    I have tried so many times but i can't really seem to understand what the question wants. Can someone please help me out in solving this question.
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*
15027
Is there an non-abelian finite group, in which more than half of the elements have order $2$   I only know that if there is one, then all elements (except identity) cannot have order $2$, otherwise it would be abelian, so there is at least one element of order $&gt;2$, say $x$. If I conjugate $x$ with another element, then conjugation preserves order but any $2$ conjugation might not give distinct elements, for the conjugacy class of an arbitrary element would be always the whole group.  Or is there such a group
24479
Show that any cyclic group of even order has exactly one element of order $2$   Attempt:  Let $G$ be a finite group of even order.  Assuming a set $A=\lbrace g \in G \vert g \neq g^{-1} \rbrace$, I can show that  $A\cup\{e\}$ has odd number of elements as $A$ has even number of elements. So $G$ has at least one element $a$ such that $a$ does not in $A$, ie $a=a^{-1}$ ie $a^2=e$. That is $G$ has atleast one element of order $2$. "More specifically, $G$ has odd number of elements of order $2$ --- ???"  But how to show that exactly one such element if $G$ be cyclic.     NB: Please don't use isomorphic properties.
86282
For which numbers $n$ exists a group of order $n$ with center {e} ?   And how many groups are there for a given order ?   The first such numbers are $6,10,12,14,18,20,21,...$ Groups of order $32,40,64$, for example have not center {e}. For $n=18$, we have $2$ groups with center {e}  Is there an easy criterion or formula ?
19687
Let $G$ be a finite group of order $n\ge2$. Which of the following statements are true? a. There always exists an injective homomorphism from $G$ into $S_n$. b. There always exists an injective homomorphism from $G$ into $S_m$ for some $m\lt n$. c. There always exists an injective homomorphism from $G$ into $GL_n(R)$.  I think b. is right (don't know how to prove though). I also think that c. is false. But not sure.
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.
49739
This is almost a duplicate of the following questions (but, read further):      But, I wasn't able to apply the proof to my case of a group of order 10 and an element of order 5.  In my case, if I try a proof by contradiction, I'm left with the fact that every element in $G$ is of order 2, but if I divide them into subgroups, every subgroup is of order 2, and has one unique element.  I also tried other methods to prove the theorem, but couldn't come up with anything meaningful.  I'm a beginner in the field, and am not familiar with theorems such as Cauchy's or Sylow's theorem (I'm aware of Lagrange's theorem). Therefore, please try to provide a simple answer, thanks!
92746
I have seen this everywhere but how do I show that : if $a$ is an element of order $n$ in a group $G$ then the order of $a^k$ is $\frac {n}{(n,k)}$.  We know, since $a$ has order $n$, then $a^n = 1$
13548
Let $G$ be a finite group having order $n$, prove that $n$ is odd if and only if for each $a\in G$, there is $b\in G$ such that $a=b^2$. I first assume $n$ is odd, then let $a$ be an element in $G$, since $a^n=e$, we have $$a=a^{n+1}=(a^{(n+1)/2})^2$$ which tells $a$ is the square of $a^{(n+1)/2}$.  To prove the converse, I use contraposition, which I assume $n$ is even, then I want to find $a\in G$ such that it is not a square. My intuition is, "assume" we can always find an element $a\in G$ such that the smallest positive integer $k$ satisfy $$a^k=e$$ is $k=n$. Then it is not true that $a$ can be expressed as a square, indeed if $a=b^2$ for some $b\in G$, then the smallest $k$ satisfy $b^k=e$ will be $k=2n$, but this contradicts the smallest $k$ cannot exceed $n$.  My problem is, is the assumption I made always true? If it is true then it should be in the very beginning of the textbook, but I have forgotton it. Any help?  Source: Apostol Analytic Number Theory.
52546
Let $G$ be a group such that the intersection of all its subgroups which are different from $\{e\}$, is a subgroup different from $\{e\}$. Prove that every element of  $G$  has finite order.  Assume that $a$ has infinite order in $G$. Then let $y\neq e$ belong to the intersection of all subgroups of $G$ different from $\{e\}$. Then $y=a^n$ and $y$ will also belong to subgroup generted by $a^2$ hence  $y=a^{2m}$ for some integers $m$. This contradicts that $a$ has infinte order.  Is this correct?
21493
Show, without Lagrange theorem, that every group $G$ of order $6$ has an element of order $3$.   I know that if $G$ has an element $a$ of order 6, then $a^2$ has order $3$, but I am not shure how to show that for the case where $G$ does not have an element of order $6$. I also know that we can only have elements $1,2,3,$ or $6$ (we proved this independent without Lagrange theorem).
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.
72182
Let $A$ and $B$ be two groups and let $G=A*B$ be their free product. Suppose both $A$ and $B$ are nontrivial. I want now to prove that $G$ is infinite and that $Z(G)$ is trivial.  I think I can prove that $G$ is infinite using this argument: two reduced words of $A*B$ are equal if and only if they have the same length and the ith letter of the first word is the same as the ith letter of the second one. Hence one can find a reduced word of $G$ of length arbitrarily big (since we are sure it won't be equal to a shorter word). Do you think it works? Has anyone another suggestion?  On the other hand, I don't really have a clue for the second part except trying to analyze a long list of cases of reduced words and prove for each one that the only element with which they commute is the identity. Does anyone have a smarter idea?  Thanks in advance!
40487
My question is:     For Which  natural numbers $n$, a finite group $G$ of order $n$ is an abelian group?   Obviouslyو for $n≤4$ and when $n$ is a prime number, we have $G$ is abelian. Can we consider any other restrictions or conditions for $n$ to have the above statement or the group itself should have certain structure as well? Thanks.
80115
I found this question in Arhangel'skii and Tkachenko's book Topological Groups and Related Structures. The first chapter of the book is devoted to algebraic preliminaries.  The question actually reads:     Give an example of an infinite abelian group all proper subgroups of which are finite.   What I have done is: Every element of this group has finite order, else we could find an infinite proper subgroup, namely the group generated by $x²$ if $x$ has infinite order.  I think this can be strengthened: every element should have a prime order. Although I haven't proved this.  Intuitively this group cannot be and infinite product of smaller groups, because you could take the product of the even group factors and find an infinite proper subgroup.  Well, this is it, a highly non-trivial problem. Thanks in advance.
115446
Find all groups of order $6$.   If there is element of order $6$ then group is cyclic. If not, I came to conclusion that there should be an element of group $G$ of order $3$ (otherwise the order of group is $2^{n},$ for some $n \in \mathbb{N} $). Let that element be $a$. Now, let $ b \in G \setminus\langle a\rangle$. It is easy to show that all elements $ e, a, a^{2}, b, ab, a^{2}b $ are different. Now I am stuck. I am sure that this isn't enough information about group $G$ but since I am a beginner in this scope I don't know what else should I write. Any hint helps!
79458
Following the advice given in , I have started to study Group Theory from the very basics. My reference text is Abstract Algebra by Dummit and Foote. While going through the exercises (page 24) I found one problem which required more effort than others:  Assume $G = \{1, a, b, c\}$ is a group of order $4$ with identity $1$. Also assume that $G$ has no elements of order $4$. Use the cancellation laws to show that there is a unique group table for $G$. Deduce that $G$ is abelian.  I proceeded with solution as follows. Since $G$ is of even order there must be at least one element of order $2$ (this statement itself was one of the problems in the exercises and is easy to handle). From the wording of the question it seems that none of the elements $a, b, c$ is singled out with specific properties and hence $a, b, c$ must behave in exactly the same manner. Hence each of them is of order $2$. Thus $a^{2} = b^{2} = c^{2} = 1$.  Next I analyze product $ab$. Clearly it can't be $a$ (as $b$ is non-identity), or $b$ (as $a$ is non-identity) or $1$ (because $ab = 1 = aa$ implies $b = a$). Hence $ab = c$ and similarly $ba = c$. By the similar nature of all elements $a, b, c$ it follows that $bc = cb = a$ and $ca = ac = b$. Thus the operation of the group is defined properly for all elements and clearly it is abelian.  I think my solution is OK, but I am not sure. Note that I have not used the fact that $G$ has no elements of order $4$. Please let me know :  1) if my solution is correct. If not then point out the flaws.  2) if there is any better / shorter solution. If so provide hints and not solution.  The question may sound too easy / simple but I request to treat me like a beginner who has never heard of Group Theory and is reading fresh from the reference text I mentioned. Thanks in advance for your inputs.  EDIT: Thinking further about this problem I wondered what would happen if an element of order $4$ was allowed in $G$. In that case I think that the group has to be isomorphic to the group $G_{1} = \{1, 2, 3, 4\}$ with modulo $5$ multiplication as the group operation. The argument is as follows. Suppose $a$ is of order $4$ and let $b$ be its inverse then $b$ is also of order $4$. Now $c$ has to be its own inverse so that $c^{2} = 1$. Again $a^{2} \neq 1$ (as $ab = 1$), $a^{2} \neq a$ (as $a \neq 1$), $a^{2} \neq b$ (as it would mean $b^{2} = 1$ and thus $b$ would be of order $2$). Hence $a^{2} = c$ and similarly $b^{2} = c$. Next we can see that $ac \neq a$, $ac \neq c$, $ac \neq 1$ (as $ab = 1$) so that $ac = b$. Similarly $ca = b$, $bc = a$, $cb = a$. Also it is clear that this group turns out to be cyclic with both $a$ as well as $b$ as generators. And it also follows that there are the only two ($G$ in original question and $G$ in new variation) groups of order $4$ upto isomorphism.  I hope I have learnt something from the answers given for the original question and this solution is correct. Please let me know if there is any problem with this reasoning.