"Let $x,y$ be elements of the group G. Prove the following is an equivalence relation: $x$ is related to $y$ iff there exists some $g$ in $G$ such that $y = g \cdot x \cdot g^{-1}$"  Now I understand that I must prove that the relation is reflexive,symmetric and transitive. But without knowledge of the set or whether the group binary operation is commutative I'm getting very confused on how to do it. I know that I just need to use the group properties of associativity, closure invertibility and identity, but I'm unsure how.  Any help appreciated.
105847
This is a question from the free Harvard online abstract algebra .  I'm posting my solutions here to get some feedback on them.  For a fuller explanation, see  post. This problem is from assignment 5.  a) Prove that the relation $x$ conjugate to $y$ in a group $G$ is an equivalence relation on $G$. b) Describe the elements $a$ whose conjugacy class (= equivalence class) consists of the element $a$ alone.  a) Let $G$ be a group and $R$ be a relation on $G$ defined by $a\sim b$ if $a$ is conjugate to $b$.  Then $a\sim b$ if there is a $g\in G$ such that $a = gbg^{-1}$.  Let $a$ be an element of $G$.  Then $a=eae^{-1}$.  So $a\sim a$ and $R$ is reflexive.  Let $a$ and $b$ be elements of $G$ such that $a\sim b$.  Then there is a $g\in G$ such that $a=gbg^{-1}$.  Then $b=g^{-1}ag$.  Since $g^{-1}\in G$, $b\sim a$.  Hence, $R$ is symmetric.  Let $a,b$, and $c$ be elements of $G$ such that $a\sim b$ and $b\sim c$. Then there are elements $g,g^\prime\in G$ such that $a=gbg^{-1}$ and $b=g^\prime cg^{\prime -1}$.  Then $a=g(g^\prime c g^{\prime -1})g^{-1}=(gg^\prime)c(g^{\prime -1}g^{-1})=(gg^\prime)c(gg^\prime)^{-1}$. Since $gg^\prime\in G$, $a\sim c$.  Hence $R$ is transitive.  Therefore $R$ is an equivalence relation on $G$. b) Let $S$ be the set of elements of $G$ such that, for $s\in S$, $s=gsg^{-1}$ for any $g\in G$.  Then $sg=gs$.  So $S$ is the set of elements that commute with every element of $G$.  In other words, $S$ is the center of $G$. Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem. Thanks.
2008
I. Let $G$ be a nonempty set closed under an associative product, which in addition satisfies:  (a) There exists an $e\in G$ such that $a\cdot e=a$ for all $a\in G$.  (b) Give $a\in G$, there exists an element $a^{-1}\in G$ such that $a\cdot a^{-1}=e$.  Prove that $G$ must be a group under this product.  II. Prove, by an example, that right indentity element and left inverse does not imply that $G$ is group.  My solution:      I. Since $G$ is closed  set under an associative product, i.e. if $a,b,c\in G$ then $(a\cdot b)\cdot c=a\cdot (b\cdot c)\in G$. Taking $c=e$ we get $(a\cdot b)\cdot  e=a\cdot (b\cdot e)=a\cdot b \in G$. We have shown that $\cdot$ is binary operation.       Since $a\in G$ then $a^{-1}\in G$ and we have the following identities $$a^{-1}=a^{-1}\cdot e=a^{-1}\cdot (a\cdot a^{-1})=(a^{-1}\cdot a)\cdot a^{-1} $$   Then    $$ \begin{align} e &amp;= a^{-1}\cdot (a^{-1})^{-1} \\ &amp;= \left( \left( a^{-1}\cdot a \right) \cdot a^{-1} \right) \cdot \left( a^{-1}\right)^{-1} \\ &amp;= \left( a^{-1}\cdot a \right) \cdot \left( a^{-1}\cdot \left( a^{-1} \right)^{-1} \right) \\ &amp;= \left( a^{-1}\cdot a \right) \cdot e \\ &amp;=a^{-1}\cdot a.  \end{align} $$   Thus we have shown that    $$ a\cdot a^{-1}=a^{-1}\cdot a=e. $$   Then we see that    $$ e\cdot a= \left( a\cdot a^{-1} \right) \cdot a = a \cdot \left( a^{-1} \cdot  a \right) = a \cdot e = a. $$       We have shown that for this set $G$ and the associative binary operation assumed to be defined on $G$, the properties of the existence of a two-sided identity element in $G$ and the existence  in $G$ of a two-sided inverse for each element of $G$ are satisfied. Therefore $G$ is indeed a group.   II. But II indeed is true. Lets take the set $G=\{a,b,e\}$ and define the product $\cdot$ by the following identities: $e\cdot e=a\cdot e=b\cdot e=e$ and $a^{-1}=b, \ b^{-1}=a$ and consider the following multiplication table for our set $G$  $\begin{array}{c | c c c c c} \hline\hline  &amp; e &amp; a &amp; b  \\ \hline e &amp; e &amp; b &amp; b &amp; \\ a &amp; a &amp; a &amp; e &amp; \\ b &amp; b &amp; e &amp; a &amp; \\ \hline  \end{array} $  It's easy to verify that conditions of second problem hold for our $G$, however, $G$ is not group since we can show that $b=a$.  Is my reasoning above correct?  EDIT: Maybe this is a duplicate but I would like to know if my solution is true since I have solved it by myself. Especiaaly I am interested in the solution of the second problem.
53649
Here, I wanted to verify that:     The property of being characteristic is a transitive relation among subgroups of a group $G$.   For subgroups $N,H\leq G$, we have $N$ char $G$ and $H$ char $N$. So $N$ char $G$ implies: $$\forall\psi\in Aut(G); \psi(N)=N$$ $H$ char $N$, so for all elements in $Aut(N)$, $H$ is remained never changing. Especially, when I take $\psi'=\psi|_{N}$ then $\psi':N\to Aut(N)$ would be in $Aut(N)$ and $\psi'(H)=H$. Since the last equality is true for $H$ and the maps, caused by restriction on $N$, then I have $H$ char $G$.  Honestly, I am inly not satisfied form the conclusion here and think I am losing something. Thanks for your hints.
40602
For all $x,y\in\mathbb{R}$   define that $x\equiv y$   if $x^{2}=y^{2}$  . Then $\equiv$  is an equivalence relation on $\mathbb{R}$  , there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements.  Solution  True. To show that \equiv  is reflexive we need to show that \forall x\in\mathbb{R}  :x=x.  Let x\in\mathbb{R}  , x\equiv\mbox{\ensuremath{x}}  if x^{2}=x^{2},   which is obvious.  [x]  ={y\in\mathbb{R}  |x\equiv y}  =[0]={y\in\mathbb{R}|0\equiv y}={0}  . Hence y^{2}  =0^{2}   =0   which implies that y=0  .
22460
I have been able to figure out the the distinct equivalence classes. Now I am having difficulties proving the relation IS an equivalence relation.   $F$ is the relation defined on $\Bbb Z$ as follows:      For all $(m, n) \in \Bbb Z^2,\ m F n \iff  4 | (m-n)$   equivalence classes: $\{-8,-4,0,4,8\}, \{-7,-3,1,5,9\} \{-6,-2,2,6,10\}$ and $\{-5, -1, 3, 7, 11\}$
7599
This problem is from Herstein's 'Topics in Algebra'. I've thought about it a bit but haven't come up with much.  Let $G$ be a non-empty set with an associative product which also satisfies:   $\exists e\in G $ such that $\forall a \in G$, $a \cdot e=a$. Given $a \in G$, $\exists y(a) \in G$, such that $y(a) \cdot a =e$.   Prove that $G$ need not be a group.  I know that $G$ is a group if any one of those multiplications is switched around, i.e, either $a \cdot y(a)=e$ or $e \cdot a=a$. But I can't quite understand why in this case $G$ is not a group. I'd be much obliged if someone can give me a good explanation of why it must be and I'd also appreciate a counter-case if nothing else.   Thanks a lot. Cheers!
51619
We have a group $G$ where $a$ is an element of $G$. Then we have a set $Z(a) = \{g\in G : ga = ag\}$ called the centralizer of $a$. If I have an $x\in Z(a)$, how do I go about proving that the inverse of $x$, $x^{-1}$, is also an element of $Z(a)$? I have already proved step 1, the subgroup test: I just need step 2, described above, and I have no idea how to start.
39789
Let $S$ be the relation on $\mathbb{N} \times \mathbb{N}$ defined by $(a,b)S(c,d)$ if and only if $ad=bc$. Prove this is an equivalence relation on $\mathbb{N} \times \mathbb{N}$.  I think I've found this to be reflexive and symmetric, but I'm stuck on transitivity. Can someone check my work so far and assist with testing transitivity?  Reflexive: Let $(x,y)S(x,y)$. Then $xy = yx$. So $S$ is reflexive. Symmetric: Suppose $(a,b)S(c,d)$. Then, $ad = bc$. Therefore, $da = cb$ and $cb = da$. Therefore, $(c,d)S(a,b)$. Thus, $S$ is symmetric.  Transitive: Suppose $(a,b)S(b,c)$. Then $ac = bb$.   This is where I'm stuck. Any ideas?
30730
Let $G$ be a group. Assume that $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ for all $a, b \in G$. Prove that the group $G$ is abelian.   Also please tell me whether there is any standard approach in proving commutativity of groups like this problem.
97815
Can someone give me an example for this question on reflexivity please. How is this reflexive? Can you show an example with numbers from the set?     Question 6   Let $X = \{0,1,2,3,\dotsc, 9\}$. Define a relation $\mathcal R$ from $X$ by "$x$ is related to $y$ if $x$ and $y$ gives the same remainder on division by $3$".   (a) Show that $\mathcal R$ is an equivalence relation.   Anwser:   (a) $\mathcal R$ is clearly both reflexive and symmetric, for every number has the same (unique) remainder on division by $3$ as itself and if $a$ has the same remainder on division by $3$ as does $b$, then $b$ has the same remainder on division by $3$ as $a$ also. Similarly $\mathcal R$ is transitive, for if $a\mathcal R b$ and $b\mathcal R c$, then $a$, $b$, and $c$ all have the same remainder on division by $3$, so $a\mathcal R c$ also.
32316
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 element, other than 0. Because if 0 is the identity element, then this group won't have inverses.  The set explicitly excludes -1, which I found to be its identity element, which makes going about proving that this is a group mighty difficult.
35719
I try to prove the following:     Let $G$ be a group and $N$ be a normal subgroup. Suppose $N$ is abelian and $G/N$ is abelian. Then $G = N \cdot (G/N)$ is abelian.   Proof:   For the first assertion that $G = N \cdot (G/N)$ let $g \in G$, then $g = g n n^{-1} \in gN \cdot N$ and then $gN \in G/N$ and $n^{-1} \in N$.  For the second assertion, let $g_1,g_2 \in G$. Then $$ g_1 g_2 = g_1 n_1^{-1} n_1 g_2 n_2^{-1} n_2 = g'_1 n_1 g'_2 n_2 $$  where $g'_i = g_i N$. Then $$g'_1 n_1 g'_2 n_2 = g'_1 g'_2 n'_2 n_2$$ since $N$ is normal, and $$  g'_1 g'_2 n'_2 n_2 = g'_2 g'_1 n'_2 n_2$$ since $G/N$ is abelian, and $$ g'_2 g'_1 n'_2 n_2 = g_2 n_2^{-1} g_1 n^{-1}_1 n'_2 n_2$$ and here I am stuck.
58037
Lets view groups as consisting of a binary operation, a distinguished element $e$, and unary operation $x \mapsto x^{-1}$. Then the group axioms can be stated as follows.   $(xy)z=x(yz).$ $xe=ex=x.$ $xx^{-1}=x^{-1}x=e.$   Supposing we dropped axiom 2 and forgot about $e$, we would need to rephrase 3. One possibility would be  3'. $xx^{-1}y=x^{-1}xy=yxx^{-1}=yx^{-1}x=y.$  So my first question is, can 3' be shortened and simplified?  Now suppose we dropped associativity, as well. Then 3' needs to be further rephrased. One possibility would be  3''. $(xx^{-1})y=(x^{-1}x)y=y(xx^{-1})=y(x^{-1}x)=y,$ and $x(x^{-1}y)=x^{-1}(xy)=(yx)x^{-1}=(yx^{-1})x=y.$  Again, my question is, can 3'' be shortened and simplified?
27790
Define $x\sim y$ means 5 divides $(x - y)$ for $x$ and $y$ integers. Show that is an equivalence relation.  Equivalence relation means it satisfies reflexity, symmetry, and transitivity.  reflexive: $x\sim x$ means 5 divides $x$  symmetry: $x \sim y \rightarrow y\sim x$ means 5 divides $x - y$ and 5 divides $y - x$: $$5/(x - y) = 5/(y - x)$$ so symmetry is satisfied.  I am not sure if I am right here and I am lost on how to prove it is transitive any suggestions would be greatly appreciated
27740
I'm stuck at this question. Can someone please help me?     Prove that if a group contains exactly one element of order 2, then that element is in the center of the group.   Let $x$ be the element of $G$ which has order 2. Let $y$ be an arbitrary element of $G$. We have to prove that $x \cdot y = y \cdot x$.  Since $x$ has order $2$, \begin{equation} x^2 = e \end{equation} That is, \begin{equation} x^{-1}=x \end{equation}  I don't really know how to proceed. I've tried a number of things, but none of them seem to work.
89491
Hei, guys!  I'm having some trouble with the next problem:  Let $A$ and $B$ be subgroups of $G$. Show that $\sim$ is an equivalence relation when it is defined as follows: $g\sim g'\Leftrightarrow g' = agb$ for some $a$ from $A$, $b$ from $B$.  Should I prove this is valid for any a and b? It says for some elements a and b. But I might as well take the unit element.   Could you give me a clue on how to solve it?  Cheers!
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.