Let $G = \mathbb R\setminus\{-1\}$. Prove that G is a group with respect to the binary operation $x*y = xy + x + y$, where $xy$ and $x + y$ denote the usual multiplication and addition of reals, respectively.
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.
15770
The following could have shown up as an exercise in a basic Abstract Algebra text, and if anyone can give me a reference, I will be most grateful.  Consider a set $X$ with an associative law of composition, not known to have an identity or inverses. Suppose that for every $g\in X$, there is a unique $x\in X$ with $gxg=g$. Show that $X$ is a group.  Note that I’m not asking for a proof (though a really short one would please me!), just some place where this has been published.
68415
I was checking my old set of homework problems that I found this one:     If $G$ is a group, show that $x^2ax=a^{-1}$ has a solution if and only   if $a$ is a cube in $G$.   One direction is easy. If $a$ is a cube in $G$ then there exists $y \in G$ such that $a=y^3$. Now with direct manipulations it is shown that $x=y^{-2}$ is a solution to $x^2ax=a^{-1}$.  The other direction is harder though. I first attempted to isolate $x$ and then somehow show that it's necessary for $a$ to be a cube. But since I had no idea how to isolate $x$ I gave up. Then I decided to prove the existence of $y \in G: a=y^3$ by showing that a particular map is bijective so I can come up with a nice variable change to conclude that $a$ must be a cube, but this attempt failed as well. I'm tired now, so I think I'm giving up. I'm looking for a hint or a complete solution if a hint is not possible at this stage.
56220
Let $G$ be a group and let $c$ be a fixed elements of $G$.    Now, I'm going to define a new operation "*" on $G$ by $a*b=ac^{-1}b$    How do I prove that the set $G$ is a group under *.  Thanks for the help!
121202
$G(x) = ax + b$ where $a$ is not equal to $0$ and $b$ are real numbers.  Prove $G$ is a one-to-one correspondence.  I understand that for every $a$ there is a corresponding $b$-value that does not repeat; however, I do not understand how I can prove this.
89132
Is it (and if yes how?) possible to define an an binary operation $*$ for an arbitrary set $M$ such that $(M,*)$ is a group?  If $M$ is finite or countable infinite this is trivial, but is it also possible for any other sets? And if yes how can you construct $*$? If no can you provide a counterexample?
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.
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.
58717
Let $G = (\mathbb{C} - \{0\}, \cdot)$, a subgroup $U = \{x+yi \mid x^2 + y^2 = 1\}$. Use the Fundamental Theorem to show that $G/(\mathbb{R}_{&gt; 0},\cdot)$ is isomorphic to $U$ where $\mathbb{R}_{&gt; 0}$ denotes the positive reals.  I am struggling to find the correct mapping from $G$ to $U$ that results in the the kernel being the nonnegative reals under multiplication. I'm hoping to get help with this mapping then I can show it is a homomorphism and onto and can use the Fundamental Theorem.
91738
Let $G$ be the group of all $n\times n$ matrices with real entries that are invertible. The operations of the group are matrix multiplication, matrix inversion, and identity matrix. Let $S$ be the set of $n\times n$ matrices with real entries and with determinant $1$. Let $R^\times$ denote the group of nonzero numbers with the operations of multiplication, multiplicative inverse, and $1$.   Prove that $G/S \cong R^{\times}$.
107447
Let $g$ be a function on $\mathbb R$ to $\mathbb R$ which is not identically zero and which satisfies the equation $g(x+y)=g(x)g(y)$ for $x$,$y$ in $\mathbb R$.   $g(0)=1$. If $a=g(1)$,then $a&gt;0$ and $g(r)&gt;a^r$ for all $r$ in $\mathbb Q$.   Show that the function is strictly increasing if $g(1)$ is greater than $1$, constant if $g(1)$ is equal to $1$ or strictly decreasing if $g(1)$ is between zero and one, when $g$ is continuous.
38710
Let $G$ be a group and $H$ a subgroup of $G$. For $x \in G$, prove $x^{-1}Hx$ is a subgroup of $G$.
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.
79201
Let $S$ be a non-empty set and $n\in \mathbb{N}, n\geq 2$ a fixed integer. We consider an associative operation $"\cdot"$ on $S$ with the following properties:   $x^{n+1}=x, \forall x\in S$ $xy^nx=yx^ny, \forall x,y\in S$   Show that this operation is commutative. This problem has been given at a romanian contest. The proof uses very complicated substitutions and I am wondering if there exists an elegant proof of this. Thank you!
81509
Let $G$ be a group s.t. every non-identity element has order 2. If $G$ is finite, prove $|G| = 2^n$ and $G \simeq C_2 \times C_2 \times\cdots\times C_2$  I know G is abelian since $ab = (ab)^{-1} = b^{-1} a^{-1} = ba$ for all non-trivial $a,b \in G$ so I have several questions remaining:    How do I prove $|G| = 2^n$? I'd like to say we use induction to prove this but I'm at a loss as to where I would start. Why is $G \simeq C_2 \times C_2 \times\cdots\times C_2$?   I've read several answers to question similar to this yet unfortunately most of them involve Galois Fields and vector spaces, both concepts I'm unfamiliar with. I'd greatly appreciate an intuitive proof.
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.
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!