Suppose that $G$ is a group with the property that $g^{2}= 1$ for all $g \in G$.Prove that $G$ is a commutactive group.  Abelian group $ab = ba$. I think like this $g*g^{-1} = 1$ after that i get stuck any hints?
121072
Let $A$ be a group, where $a^2=1$ and $a$ belongs to $A$. Prove that this group is commutative.  Thank you for help.
108895
Let $G$ be a group and $A$ &amp; $B$ be two sets s.t. $G$ acts transitively on each of $A$ &amp; $B$. Choose some $\alpha$ and $\beta$ in $A$ &amp; $B$ respectively then prove that if $G=G_\alpha G_\beta$ then $G=G_x G_y$ for all $x$ $\in A$ &amp; $y$ $\in B$   where $G_x$ is stabilizer of $x$ in $G$ and similarly $G_\alpha$ be the stabilizer of $\alpha$ and $G_y$ is the stabilizer of $y$.  I know we have to choose some elements and apply some manipulative trick. it is hardly 2 liner but just can't think of right elements. help!
111798
If $G$ is a group of order $6$. How do I show that there is some $x\in G$ with $x^2\neq e$?. I was thinking of trying a contra positive approach, by proving that if there is some $x^2=e$ then $G$ is Abelian. But how would I continue after showing that $x^2=e$?
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?
16646
Suppose $a$ and $b$ are elements of a group $G$. If $a^{-1}b^{2}a=b^{3}$ and $b^{-1}a^{2}b=a^{3}$, prove $a=e=b$.   I've been trying to prove but still inconclusive. Please prove to me.  Thanks very much for proof.
3922
G is the set of matrices of the form $G=$$\begin{pmatrix} x &amp; x \\ x &amp; x \end{pmatrix}$. So for this set to be a group I know it needs to be:    Closed under matrix multiplication The Associative Property holds Contains an Identity Element Every element needs to have an inverse   So the form of the matrices is such that all the elements are the same but not 0. How do I go about proving these?  Working through this problem, I seem to have hit a contradiction. Since G is a subgroup of the bigger $2x2$ nonsingular matrices group why does G not have the same identity element as its parent group? Namely \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 1 \end{bmatrix}  Isn't the subgroup supposed to have the same identity element as its parent group?
44724
Let $G$ be a group, and suppose that $a= {a^{-1}}$ for every $a\in G$. Prove that $G$ is Abelian.    I know that I need to prove that ${a^{-1}}$ = a and use right multiplication. I also know that I need to declare a, b $\in$ G and go from there, but I'm confused when I get to   (${a^{-1}}$${b^{-1}}$) = ab.
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.
32630
Well, this is an exercise problem from Herstein which sounds difficult:   How does one prove that if $|G|&gt;2$, then $G$ has non-trivial automorphism?   The only thing I know which connects a group with its automorphism is the theorem, $$G/Z(G) \cong \mathcal{I}(G)$$ where $\mathcal{I}(G)$ denotes the Inner- Automorphism group of $G$. So for a group with $Z(G)=(e)$, we can conclude that it has a non-trivial automorphism, but what about groups with center?
107116
I want to prove for a group $G$, that if  $$a\circ b =a\circ c$$ then this is true $$b=c$$ I started with $b=b\circ e$, but this didn't help me at all.   Next I tried with this: $$(a\circ b)\circ c=a\circ (b\circ c)$$ but I don't know/understand how to go further. How can I prove this equation?
44474
Let $G$ is a group and $a, b \in G$. If $a$ has order $6$, then the order of $bab^{-1}$ is...  How to find this answer? Sorry for my bad question, but I need this for my study.
83231
Let $A$ be a normal subgroup of a group $G$. Let $B$ be a subgroup of   $A$ with the property that whenever $B_1 \leq A$ is conjugate to $B$   in $G$ then it is conjugate to $B$ in $A$. Prove that $G=AN_G(B)$   Help me some hints.
38649
Let $G$ be a group and $g,h \in G$. I need to prove that if $g$ and $h$ commute and their orders are coprime, then $|gh| = |g||h|$, that is, the order of their product is the multiple of their orders.  Since $gh = hg$, then $(gh)^{ lcm(|g|, |h|)} = e$, so, $|gh|$ divides $lcm(|g|, |h|)$. Since $\gcd(|g|, |h|) = 1$, then $lcm(|g|, |h|) = |g||h|$.  So, $|gh|$ divides $|g||h|$.
56191
Let $H = \{x \in G: x = y^2$ for some $y\in G\}$;  that is, let $H$ be the set of all the elements of $G$ which have a square root.  Prove that $H$ is a subgroup of $G$, where $G$ is an abelian group.  So I'm pretty lost in group theory.  I know that I need to prove that for any element, $a,b \in H$, that $(a^{-1} * b) \in H$, where * is the group operation on $G$.   I know that I also need to use the fact that $G$ is abelian.  So somehow commutativity is involved.  But I admit that I have no idea what I am doing.  I guess my primary question is:  What does $x^{-1}$ look like in $H$?  Since $x$ = $y^2$ then $x^{-1} = (y^2)^{-1}$, which is what?
27569
I am looking for an example of a group $G$ where the equality $(ab)^{n}=a^{n}b^{n}$ holds for two consecutive integers $n$, but $G$ is not an abelian group. I've started do some calculations in the group $D_{4}$ (I gave up!) Do you know where I can find that example? Is it possible find such example without doing a lot of calculations?  Thanks for your help!
123050
If $\forall x \in G$ where G is a group $x=x^{-1}$ then how is this group abelian .   This question is from Gallian's book where question provides information that G is a group and $\forall x \in G$ , $x=x^{-1}$ then we have to prove that G is abelian , how is it possible ??
2480
Let $G$ be a finite semi-group with identity such that it has only one idempotent.Is $G$ a group?  It only remains to show that for any $a\in G$ $\exists b\in G$ such that $ab=ba=e$ where $e$ is the identity of $G$  Also $e$ is the only idempotent of $G$ .How to proceed next?