How can you prove that if each element of group is inverse to itself then the group is commutative?
4434
Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.   This question is from group theory in Abstract Algebra and no matter how many times my lecturer teaches it for some reason I can't seem to crack it.   (Please note that $e$ in the question is the group's identity.)  Here's my attempt though...  First I understand Abelian means that if $g_1$ and $g_2$ are elements of a group $G$ then they are Abelian if $g_1g_2=g_2g_1$...  So, I begin by trying to play around with the elements of the group based on their definition...  $$(g_2g_1)^r=e$$  $$(g_2g_1g_2g_2^{-1})^r=e$$ $$(g_2g_1g_2g_2^{-1}g_2g_1g_2g_2^{-1}...g_2g_1g_2g_2^{-1})=e$$  I assume that the $g_2^{-1}$'s and the $g_2$'s cancel out so that we end up with something like,  $$g_2(g_1g_2)^rg_2^{-1}=e$$ $$g_2^{-1}g_2(g_1g_2)^r=g_2^{-1}g_2$$  Then ultimately...  $$g_1g_2=e$$  I figure this is the answer. But I'm not totally sure. I always feel like I do too much in the pursuit of an answer when there's a simpler way.  Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra.
80308
I am tasked to show that the set of affine transformations of ,    Where S is an invertible linear transformation and b is a vector, forms a group under the composition of maps.   It was easy to show that   Now I need to show that there is an identity element, and that all elements are invertible such that   Im a little unsure what to do at this point....Can anyone help me on this? Thanks!
32393
Let $R$ be a ring satisfying: $\forall x\in R, \; x^3=x$. Prove that $R$ is a commutative ring.
33068
Let $(G,*)$ be a group with identity $e$ , let $a,b∈G$ such that $a*b^3*a^{-1}=b^2$ and $b^{-1}*a^2*b=a^3$ , then how do we prove that $a=b=e$ ?
82712
Is my proof ok?  Let $f:G\to G^{\prime}$ be a group homomorphism and let $H\lt G$. $Im(H) = \{f(x):x\in H\}$. To show that $Im(H)$ is a group, it suffices to show that $f(x)f(y)^{-1}\in Im(H)$. $$f(x)f(y)^{-1}=f(xy^{-1})\in Im(H)\text{ because } xy^{-1}\in H$$  $Ker(f)=\{x:f(x)=e^{\prime}\}$ ($e^{\prime}$ is the identity in $G^{\prime}$). $$f(xy^{-1})=f(x)f(y)^{-1}=e^{\prime}(e^{\prime})^{-1}=e^{\prime}\in Ker(f)$$.
38710
Let $G$ be a group and $H$ a subgroup of $G$. For $x \in G$, prove $x^{-1}Hx$ is a subgroup of $G$.
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 $|g||h|$.
111263
I need to proof for a finite abelian group $G$ that for all $x\in G$ we have $\prod_{g\in G}xg=\prod_{g\in G}g$.  I figured that using the commutative property $\prod_{g\in G}xg=x^n\prod_{g\in G}g$. Which would leave us to proof $x^n=e$, where $e$ is the identity element.  Hopefully I'm just missing something obvious here, because I feel like smashing my head against the table now.
30991
If $x^{2}-x\in Z(R)$ for all $x\in R$, then $R$ is commutative.  I need to proof this theorem and I have something like this below. However, I do not know how to continue this proof.   $(x+y)^{2}-(x+y)=x^{2}+xy+yx+y^{2}-x-y=(x^{2}-x)+(y^{2}-y)+xy+yx\in Z(R)$  so $xy+yx\in Z(R)$
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.
24596
I am trying to prove that $(G, *)$ is an abelian group with $G=(-1,1)$ and $a*b=$$\frac{x+y}{1+xy}$.  Thus far I have found that the identity element $e=0$. From here, I set $a*b=0$ and found $a^{-1}$ to be $-a$.   My work for trying to prove closure and that the set is abelian is:  Let $a,b \in G$, where $a=e$ and $b=a^{-1}$. Does  $a*b=b*a$?  Evaluating, I see that $a^{-1}=a^{-1}$ which then proves that G is closed and G is an abelian group.  Did I go about this proof correctly and efficiently?
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.
100824
M is a commutative monoid with identity element denoted by e. U(M) is the set of all invertible elements of M. Let a∈M and b∈M. Prove that if ab∈U(M), then a∈U(M).  I'm not sure if the way I approached this question is legitimate or whether there's a clearer way to show it. Method: Let ab ∈ U(M). Hence $(ab)^{-1}$ ∈ U(M). ⇒ $b^{-1}a^{-1}$ ∈ U(M) ⇒ (ab)($b^{-1}a^{-1}$) ∈ U(M) ⇒ $aa^{-1}$ ∈ U(M) Similarly for ba ∈ M we have: $(ba)^{-1}$ ∈ U(M) ⇒ $a^{-1}b^{-1}$ ∈ U(M) ⇒ $a^{-1}b^{-1}$(ba) ∈ U(M) ⇒ $a^{-1}$a ∈ U(M). As a ∈ M has a right inverse and left inverse in U(M), we can conclude that a ∈ U(M).
42552
Let $a$ be an element of a group G. Prove that $a$ commutes with each of its conjugates in $G$ iff $a$ belongs to an abelian normal subgroup of $G$.   My Try:  I proved the backward direction. For forward one, given $g\in G, a(gag^{-1})=(gag^{-1})a$. I was trying to prove that $a\in Z(G)$. In order to prove that, I must show that $ag=ga$. But failed. Am I on the right track? Can anybody please give me a hint? I want to try it myself.
62301
Let $G$ be a set with a binary operation *, associating to each pair of elements $x$ and $y$ of $G$ a third element $x*y$ of $G$. Suppose that the following properties are satisfied:   $(x*y)*z = x*(y*z)$ for all elements $x$, $y$, and $z$ of $G$ (the Associative Law); there exists an element $e$ of $G$ such that $e*x = x$ for all elements $x$ of $G$; for each element $x$ of $G$ there exists an element $x'$ of $G$ satisfying $x'*x = e$.   How to prove  i. $x*e=x$ and  ii. $x*x'=e$,  so that $G$ is a group?
87274
In a course of group theory an exercise is left with no indications :      Let $A$ be an abelian group of finite type and $a_1,\dots ,a_n$ a $\mathbb{Z}$-basis of $A$. Let $f : A \to A$ be a group homomorphism. Les $M=(m_{ij})$ be the matrix (invertible) defined by $\displaystyle f(a_j)=\sum_{i=1}^n m_{ij}a_i$ with $\det (M)\neq 0$. Show that $[A:\operatorname{Im}(f)]=|\det M |$, i.e the index of $\operatorname{Im}(f)$ in $A$ as a subgroup is precisely $\det (M)$.   I am not able to prove this fact, and need a bit of help. In fact I do not see where I can make $\det (M)$ appear. Can you give me some hints ?   Thank you
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.