Let $G$ be a group and $G'=\langle [x,y] : x , y \in G\rangle $, where $[x,y]=x^{-1}y^{-1}xy$.  I am trying to prove that $G/G'$ is an abelian group.  What I've done: $$X\in G/G' \Rightarrow X=xG' ; x\in G$$ $$Y\in G/G' \Rightarrow Y=yG' ; y\in G$$ $$XY=(xG')(yG')=xyG'$$ $$YX=(yG')(xG')=yxG'$$  Now, I have to prove that $XY=YX$, but I don't know how to do that :( Any help would be appreciated.
57349
Trying to get my head around the commutator subgroup. This is an excercise from Artin's Algebra:  Let $C$ be the commutator subgroup of $G$. Prove that $G/C$ is abelian.  Here is what I've done:  Let $xC,yC \in G/C$ then $xyx^{-1}y^{-1}C = C$ since $xyx^{-1}y^{-1}$ is a commutator hence belongs to $C$. But then $xyC = yxC$ so $xC$ and $yC$ commute in $G/C$. This can be done for any elements, so $G/C$ is abelian.  This seems somewhat surprisingly short. Is that all there is to it?   Regards
71010
Let $G$ be the group  of all $2 \times 2$ matrices of the form $\begin{pmatrix} a &amp; b \\ 0 &amp; d\end{pmatrix}$ where $ad \neq 0$ under matrix multiplication. Let  $N=\left\{A \in G \; \colon \; A = \begin{pmatrix}1 &amp; b \\ 0 &amp; 1\end{pmatrix} \right\}$ be  a subset  of the group $G$. Prove that  $N$ is  a normal subgroup of $G$ and prove that  $G/N$ is abelian group.   Here is my attempt!  To prove $N$ is normal I consider  the group homomorphism $f \colon G \to \mathbb R^*$ given by $f(B) = \det(B)$ for all $B$ in $G$. And I see that $f(N)$ is all the singleton $\{1\}$ since $\{1\}$ as a subgroup of $\mathbb R^*$  is normal, it follows that $N$ is also normal. Is this proof helpful  here? Then how to prove that $G/N$ is Abelian? I know $G/N$ is a collection of left cosets.   Thank you.
85573
if G is generated by {a,b} and ab=ba, then prove G is Abelian  All elements of G will be of the form $a^kb^j,\,\, k,j \in \mathbb{Z}^+$   so I need to get $a^kb^j = b^ja^k$
44724
Let $G$ be a group, and suppose that $a= {a^{-1}}$ for every $a\in G$. Prove that $G$ is Abelian.    I know that I need to prove that ${a^{-1}}$ = a and use right multiplication. I also know that I need to declare a, b $\in$ G and go from there, but I'm confused when I get to   (${a^{-1}}$${b^{-1}}$) = ab.
72375
Let G be a set of all non-zero real numbers. And  a * b  = (ab/2).  We need to show (G,*) is an abelian group.  I know I need to show ab = ba, but that seems so trivial,  Assuming the group is abelian,  a * b  = b* a  ab/2 = ba/2 Crossing out 2, we get ab = ba  Is this the correct answer ?  Multiplying both sides inverse of a and b results in 1 = 1 . Am I doing something wrong ? Or am I missing something?
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Let $\langle g\rangle$ be a cyclic group of order $n$. Suppose $1\leq q \leq n-1$, I want to show that $g^q$ generates $\langle g\rangle$ if and only if $\gcd(n,q)=1$.  Suppose $g^q$ generates $\langle g\rangle$, then   $$ 1,g^q,g^{2q},...,g^{(n-1)q} $$  are all distinct and $g^{nq}=1$, since $n$ is the order of the group. Note that this is the lowest multiple of $q$ for which this is the case. Therefore $\operatorname{lcm}(q,n)=qn$ and it follows that $q$ and $n$ are coprime.  On the other hand, if $\gcd(q,n)=1$, then $\operatorname{lcm}(q,n)=qn$, we have $g^{nq}=1$ and hence all of  $$ 1,g^q,g^{2q},...,g^{(n-1)q} $$  are distinct again and $g^q$ is a generator.  Is my proof correct at all and is there maybe a more elegant argument?
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Suppose $G$ is a group generated by elements $x$ and $y$ where $xy^2 = y^3x$ and $yx^3 = x^2y$ What can you prove about $G$?  I've just been playing around with the relations but I can't seem to get anywhere. It seems as if I'm going in circles. I don't know what to look for in terms of how the equations should look in order to give me an insight into the structure of the group.  So given a group presentation, are there any general strategies of what to look for in the relations (as in how to rearrange them) in order to obtain insight about the group?
51751
Possible Duplicate:             If $G$ is a group and $Z(G)$ the center of $G$, show that if $G/Z(G)$ is cyclic, then $G$ is abelian.  This is what I have so far:  We know that all cyclic groups are abelian. This means $G/Z(G)$ is abelian.  $Z(G)= \{z \in G \mid zx=xz \text{ for all } x \in G \}$.  So $Z(G)$ is abelian.    Is it sufficient to say that since $G/Z(G)$ and $Z(G)$ are both abelian, $G$ must be abelian?
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.
100999
If $H \leq Z(G) \leq G$, where G is a finite group, $Z(G)$ is its center, and $(G:H) = p$ for some prime $p$, then G is abelian.  Well because H &lt; Z(G) we have that H commutes with all of G. Also |G/H|=p. I was thinking of connecting this result to the fact that if $G/Z(G)$ is cyclic, then $G$ is abelian, but was unable to. Then I thought about using the result that $(G/H)/(Z(G)/H) = G/Z(G),$ but that leads to nowhere. Any suggestions?
93811
Suppose that $G$ is a group with the property that for every choice of elements in $G$, $axb=cxd$ implies $ab=cd$. Prove that $G$ is Abelian. (Middle cancellation implies commutativity).  I am having trouble with this homework problem. The way I started was:  Suppose $G$ is a group for which middle cancellation holds. Multiply $x$ by its inverse, $x^{-1}$ using middle cancellation so you have $ab=cd$. Thus $G$ is Abelian.   I am unsure if this is the way to approach it.
69006
Possible Duplicate:             Let $(G,\star)$ be a group with identity element $e$ such that $a \star a = e$ for all $a \in G$. Prove that $G$ is abelian.   Ok, what i got is this: we want to prove that ab=ba, i.e. if aa=e , a=a' where a' is the inverse and bb=e, b=b' where b' is the inverse so ab=(ab)'=b'a'=ba....
88248
Let $N$ be a normal subgroup of a group $G$ such that every subgroup of $N$ is normal in $G$ and $C_G(N)\subseteq N $. Prove that $G/N$ is abelian.   Here, as usual, $C_G\left(N\right)$ means the centralizer of $N$ in $G$ (i.e., those elements of $G$ that commute with everything in $N$).  I think we need to use that every subgroup of $N$ is normal in $G$ but i can't use .Please help me with Hints.
3922
G is the set of matrices of the form $G=$$\begin{pmatrix} x &amp; x \\ x &amp; x \end{pmatrix}$. So for this set to be a group I know it needs to be:    Closed under matrix multiplication The Associative Property holds Contains an Identity Element Every element needs to have an inverse   So the form of the matrices is such that all the elements are the same but not 0. How do I go about proving these?  Working through this problem, I seem to have hit a contradiction. Since G is a subgroup of the bigger $2x2$ nonsingular matrices group why does G not have the same identity element as its parent group? Namely \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 1 \end{bmatrix}  Isn't the subgroup supposed to have the same identity element as its parent group?
41419
$G$ is a group and $f:G \rightarrow G$ is a function defined as $f(a)=a^{-1}$ where $a^{-1}$ is the inverse of $a$ under the group operation. Prove that $f$ is an isomorphism if and only if $G$ is abelian.    I understand that I have to prove $f(ab)=(ab)^{-1}=b^{-1}a^{-1}$. How might I do that?  Reference: Fraleigh p. 49 Question 4.40 in A First Course in Abstract Algebra
94188
Let $G$ be a group. If $G / Z(G)$ nilpotent, then prove that $G$ is nilpotent.   The way I have worked out this question is fairly tedious, and I would be interested if there is a more straightforward approach here.  I solved this problem by showing that if $G / Z(G)$ is of nilpotence class $k$, then $G / Z(G) = Z_{k}(G / Z(G)) = Z_{k+1}(G) / Z(G)$, the latter equality shown using a tedious induction proof. Then the conclusion readily follows.  Is there a different approach to this problem, or is there a short way to prove that $Z_{k}(G / Z(G)) = Z_{k+1}(G) / Z(G)$?
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Let $G$ be a group and let $a \in G$ be a fixed element. Let $$H= \{\text{$g \in G$ | $g^{-1}ag=a$}\}.$$  Prove that $H$ is a subgroup of $G$.    So I know I need to show:  (i) $H$ is closed under the operation on $G$.  (ii) $H$ is closed under inversion.  (iii) $H\neq \emptyset$    For closure:  Suppose $x,y \in H.$ Then $x^{-1}ax=a$ and $y^{-1}ay=a.$ So  $$\begin{align} (xy)^{-1}a(xy)&amp; = y^{-1}x^{-1}axy \\  &amp; = y^{-1}(x^{-1}ax)y \\  &amp; = y^{-1}ay \\   &amp; = a \\ \end{align}$$ Thus $xy \in H$.    Now how do I show it's closed under inversion? I'm not sure what the inverse is...   And to show $H \neq \emptyset$ all I need to do is show that the identity is in the set. And I'm not sure how to do that either.  Help!