I need to show that the following is an abelian group:     $$x*y = \frac{x+y}{xy+1}$$ on the set $\{x \in \Bbb R \,|\, -1 &lt; x &lt; 1\}$.   I have been working on this problem, trying to show closure. I know that we need to show that $|x+y|&lt;|xy+1|$ for all $x, y \in (-1,1)$. Can I assume that the max value that the expression yields is $1$ if we take $x=1$ and $y=1$? And the lowest value that is possible is when $x=-1$ and $y=1$? Or am I going about this the wrong way?
122977
I have to prove that the result of $x * y \in G$ so $\frac{x+y}{1+xy} \in (-1, 1)$. So $x &gt; -1$ and $y &gt; -1$ at the same time $x &lt; 1$ and $y &lt; 1$.  If I multiply the first 2 expressions I obtain $xy &lt; 1$ which is true, and if I sum up the last 2 I get $x + y &lt; 2$. At this point I am not sure if I am doing well. How should I continue this problem?
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 $e$.
51653
Did not find this from this website...  If $$ \gcd(a,b)=1,$$ then there exists integers $x$ and $y$ such that $$xa+yb=1.$$  Now, the tip is to use particular corollary, that states:  The class $[m]_{n}$ generates $\mathbb{Z}/n\mathbb{Z}\Leftrightarrow \gcd(m,n)=1.$  I am totally lost with the corollary.  Let's assume that $\gcd(m,n)=1$. Then, $[m]_{n}$ generates $\mathbb{Z}/n\mathbb{Z}$.  OK! Then what?  There is also follow up, where I have to prove the converse. I am familiar with Bezout's lemma.
93658
Suppose $a$ and $z$ are in $\mathbb{C}$ with $|z|=1$ and $|a|&lt;1$. I want to show $$\left|\frac{z-a}{1-\bar{a}z}\right|=1$$ but I'm stuck on dealing with $a$. Is this just another complex number of the form $a+bi$?   How do I show this is true? Thanks in advance.
90771
Let $G$ be an finite abelian group, define $s= \Pi_{g\in G} g$, show that $s = e$ unless $G$ has exactly one involution $u$, in this case $s=u$.   I have the start of a proof, but I get stuck at a point.   Proof  Write $G$ as $G = \{e\} \cup I \cup \{g_1, g_1^{-1}, \ldots, g_n, g_n^{-1}\}$ where $I$ is a set of the all the groups involutions.  Since $G$ is abelian $\Pi_{g\in \{g_1, g_1^{-1}, \ldots\}} g = e$ we just need to examine $\Pi_{u\in I} u$  Now:   If $|I|=0$ then the theorem holds, If $|I| = 1$ then $I=\{u\}$ and $s= u$, the theorem holds If $|I| = 2$, requires closer examination:  Let $I = \{u_1,u_2\}$ be the set of all involutions (not equal to $e$), then $u_1u_2$ is also an involution (since $G$ is abelian).   If $u_1u_2 =e$ then $u_1=u_2$, contradiction, if $u_1u_2 = u_1$ then $u_2=e$, contradiction. $u_1 u_2=u_2$ results in the same contradiction. This means that the case $|I|=2$ is not possible. the case $|I|=3$ is possible, let $I = \{u_1,u_2,u_3\}$ then it's easy to prove $u_3 = u_1u_2$. The theorem holds.   How should I continue? I've tried splitting $I$ as   $$I=\{u_i: u_i \not = u_ju_k (\text{for a certain }j\not = k)\} \cup \{u_i = u_ju_k (\text{for a certain} j \not = k)\}$$ but I still get stuck.  Also the case $|I|= 4$ seems contradictory once again..
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
78108
I need some help on this one.     $G$ is a group. If $N$ is a subgroup of $G$ contained in the centre of $G$ and $G/N$ is cyclic, show that $G$ is Abelian.   My attempt is only half way and stuck at the bit at the end...     If $G$ can be shown to be cyclic, then it is Abelian, since $G=&lt;g&gt;$ then $g_1,g_2 \in G$ are $g_1=g^n,g_2=g^m$ for some $n,m$ and $g_1g_2=g^ng^m=g^{n+m}=g^{m+n}=g_2g_1$.       By definition of a centre of $G$, $\forall n \in N$, $gn=ng$ for any element in $G$. Namely, $N$ must be a normal subgroup.   (Here, I pondered if I should just proceed or change gears to looking at homomorphisms of $G$ since $N$ is normal).      Now $G/N$ is cyclic so $G/N=&lt;gN&gt;$ for some $g \in G$. Then, $\forall g'N \in G/N$ there is some $m$ so that $(gN)^m=g^mN=g'N$. Therefore, for the representative elements of $G/N$, I have $g^m$.   Here I get stuck. So if $g'N$ then I have that each $g'$ is cyclic but as an equivalent class, I cannot guarantee that all $g'$ can be represented in the form $g^m$. The only thing I can guess is the canonical map $p:G \to G/N$ that might be relevant to solving this but again, I don't know how specifically it would help.  Does anyone know how to solve this?
12693
Is the following statement true or false:  If $G$ is a group with the property that $g=g^{-1}$ for all $g \in G$, then $G$ is abelian.   I believe it is false since I know that abelian or commutative property implies that every element in $G$ must have an inverse. Thus $g\cdot g^{-1}= g^{1}\cdot g = e$. I need someone to check my attempt.
53446
Could you give me a simple example of $G$ abelian with ${\rm Aut}(G)$ non-abelian? Otherwise how could I prove that $G$ abelian implies ${\rm Aut}(G)$ abelian. (I don't really think that's true)
119189
I have a question;   X and Y are subgroups of a group G.  If $|X|, |Y| &lt; \infty$, show that   $|XY| = \frac{|X||Y|}{|X \cap Y|}$  but I can't really picture what it is talking about to even get started.  Is it to do with looking for different ways so combine the elements {x, y, xy} ?  Thanks
15672
Suppose $G=\{a_1,...,a_n\}$ is a finite abelian group, and let $x=a_1a_2\dotsm a_n$.   Prove that if there is more than one element of order $2$ then $x=e$.   What I've done so far: (#1 is just for illustration, There should be at least $3$ such elements)   $G$ has an even number of elements with order $&gt;2$ (They are paired so: $(a,a^{-1})$ ). And then there are $e$ and $\{b|o(b)=2\}$. Also, if $G$ has at least one element of order 2 then $|G|$ is even (Lagrange). Thus $|\{b|o(b)=2\}|$ must be odd. If there is exactly one element $b$ of order $2$ then $x=b$. That's because $G$ is abelian, and we can write: $x=e\cdot b\cdot(a_1 a_{1}^{-1}\dotsm a_n a_{n}^{-1})=b\cdot(e\dotsm e)=b$. If there are exactly $3$ such elements $a,b,c\in G$, then, as shown above, $x=abc$ and thus $x=e$ (That's because $(abc)^2=a^2b^2c^2=e$ which means $abc\in \{a,b,c,e\}$. If $abc=a$ then $b=c^{-1}=c$ which is false, same goes for $abc=b,abc=c$.)   I haven't managed to find a reason why the claim must hold for $5, 7, ...$ (i.e. All other odd integers). Does anyone have an idea?  Thanks in advance. Please excuse my English.  edit: Not a canonical answer but a simple one (if such even exists).
93764
Let n be a positive integer and let G be an Abelian group. Define  $H := \{x ∈ G : |x| \ \text{divides} \ n\}$  where x denotes the order of |x|.  Show that H is a subgroup of G.  I know I need to prove that H is non-empty, contains the identity, e, and that it contains $ab^{-1}$.  I'm just not sure how to prove this, thank you for any help!
36424
While reading Cours d'algèbre by D. Perrin, I found the following claim:     Proposition. If $G$ is a group, then $D(G)\subseteq G^2$, where $G^2$ is the subgroup generated by the squares of $G$.   I understand that it suffices to prove that for all $x,y\in G$, $[x,y]:=xyx^{-1}y^{-1}$ is a product of squares.  Equivalently, since $G^2$ is characteristic, hence normal, it suffices to establish that $G/G^2$ is abelian.  For now, none of my attempts have been fruitful. Any enlightenment will be greatly appreciated!
122407
I want to show that $\forall x$,$y\in G$, $(xy)^2=x^2y^2 \iff G $ is an abelian group.  Proof of the backward direction seems easy enough:  Assume $G$ is an abelian group. Then consider $(xy)^2=(xy)(xy)$. By associativity, $(xy)(xy)=xyxy$. Since $G$ is abelian, $xyxy=xxyy=x^2y^2$.   Proof of the forward direction I attempt by contrapositive:  That is, assume $G$ is not an abelian group. Then, consider $(xy)^2=xyxy$ as above. Yet, $xyxy\neq xxyy=x^2y^2$ since we no longer have commutativity.  Is this proof by contrapositive completely correct?
84244
If $G$ is a group in which $(a\cdot b)^{i} =a^i\cdot b^i$ for three consecutive integers $i$ for all $a, b \in G$, show that $G$ is abelian.  Proof: Let $x$ be the smallest of the 3 consecutive integers. Then, we have   $(1)(a\cdot b)^{x} =a^x\cdot b^x$, $(2)(a\cdot b)^{x+1} =a^{x+1}\cdot b^{x+1}$ and $(3)(a\cdot b)^{x+2} =a^{x+2}\cdot b^{x+2}$.  Using $(2)$ and multiplying $a^{-1}$ on the left and $b^{-1}$ on the right, we get $baba...ba = a^{x}b^{x}(4)$ whereby there are $x$ number of $a$ and $x$ number of $b$ on both sides.  Using $(1)$ and multiplying both sides on the right by $ab$, we have $\overbrace{abab...ab}^{(x+1)ab} = (a^{x}b^{x})ab (5)$.  Substitute $(4)$ into $(5)$, we get $\overbrace{ab...ab}^{(x+1)ab} =\overbrace{ba...ba}^{(x)ba}ab(6)$.  Using $(3)$ and multiplying $a^{-1}$ on the left and $b^{-1}$ on the right, we get $\overbrace{ba...ba}^{(x+1)ba} = a^{x+1}b^{x+1} = \overbrace{abab...ab}^{(x+1)ab}(7)$.  Combining $(6)$ and $(7)$, and multiply $a^{-1}b^{-1}...a^{-1}b^{-1}$ on the left, we get $ab = ba$. Hence $G$ is abelian.
43191
Let $x,y,z\in\Bbb R$, and at least one is be a postive integer, and such that  $$x+\frac{1}{y},\;y+\frac{1}{z},\;z+\frac{1}{x}\in\Bbb Z$$ Find the value of $x,y,z$.  My try:  $$\left(x+\dfrac{1}{y}\right)\left(y+\dfrac{1}{z}\right)\left(z+\dfrac{1}{x}\right)=\dfrac{(xy+1)(yz+1)(xz+1)}{xyz}\in Z$$  Then I can't. Thank you.  I konw $(x,y,z)=(1,1,1),(-1,-1,1),(2,0.5,1)$ is such condition,But How prove it?
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?