Linked Questions

3
votes
6answers
3k views

How to prove that if each element of group is inverse to itself then group commutative? [duplicate]

How can you prove that if each element of group is inverse to itself then the group is commutative?
1
vote
2answers
7k views

Let $G$ be a group in which $a^2=e$ for all elements of $a$ of $G$. Show that $G$ is Abelian. [duplicate]

Let $G$ be a group in which $a^2=e$ for all elements of $a$ of $G$. Show that $G$ is Abelian. I need help on this problem. Appreciated!
3
votes
5answers
511 views

Prove that a group where $a^2=e$ for all $a$ is commutative [duplicate]

Defining a group $(G,*)$ where $a^2=e$ with $e$ denoting the identity class.... I am to prove that this group is commutative. To begin doing that, I want to understand what exactly the power of 2 ...
4
votes
3answers
2k views

Proving G is commutative. [duplicate]

Suppose that $g^2=e$ for all elements $g$ of a group $G$. Prove that $G$ is commutative. How would I go about doing this proof? I understand what it means by $g^2=e$, and a group.
2
votes
5answers
223 views

Let $A$ be a group, where $a^2=1$, a belongs to $A$. Prove that this group is commutative. [duplicate]

Let $A$ be a group, where $a^2=1$ and $a$ belongs to $A$. Prove that this group is commutative. Thank you for help.
7
votes
2answers
390 views

If every nonidentity element in a group is of order $2$, the group is abelian [duplicate]

Let $G$ be a group. Show that if every non-identity element in $G$ has order $2$ then $G$ is abelian. Proof: Let $a,b $ be non-identity elements in $G$. Since $|a|=|b|=2$ , that means $ab=babaab$ $=$...
1
vote
3answers
339 views

Prove that a group G such that every element $g \in G$ satisfies the equality $g^2 = 1_G$ is abelian [duplicate]

Possible Duplicate: Prove that if $g^2=e$ for all g in G then G is Abelian. This is how I proved it: Abelian means that the following axioms hold: Associativity, Existence of Identity and ...
1
vote
2answers
205 views

If $g^2 = e$ for all $g \in G$, then $G$ is abelian [duplicate]

Let $G$ be a group. Prove that $g^2 = e$ for all $g \in G$, then $G$ is abelian. ($e$ is the identity element.) My Solution: Let $a,b \in G$. Then $a(ab)b = a^2b^2 = e^2 =e$. Now I tried to reverse $...
1
vote
3answers
119 views

Showing a group is abelian [duplicate]

Suppose that $e$ is the identity in a group G and that $a^2 = e $ for all $a\in G$. Show that G is abelian. How do I figure out what the binary operation is to show that G is abelian i.e. if this is ...
2
votes
2answers
312 views

Let $G$ be a group. Prove that IF $x^2 = e$ for all $x \in G$, then $G$ is abelian. [duplicate]

Let $G$ be a group. Prove that IF $x^2 = e$ for all $x \in G$, then $G$ is abelian. My attempt: $x^2 = e$ $x = x^{-1}$ $xx^{-1}=x^{-1}x^{-1}$ $e = x^{-1}x^{-1}$ $e = (xx)^{-1}$ $e = (x^2)^{-1}$ ...
2
votes
4answers
118 views

Let $(G,\circ)$ be a group in which $x\circ x$ is the neutral element for all $x \in G$ [duplicate]

Prove that $x=x^{-1}$ for all $x \in G$ and that G is commutative. I have no idea where to start. I know that commutative will mean that for $x,y \in G$ $x\circ y = y \circ x$ but I don't know ...
1
vote
1answer
40 views

A monoid where square of all elements are 1 is abelian [duplicate]

The following problem gives me a very hard time: Let $M$ be a monoid with $a^2 = 1$ for $a \in M$. Show that $M$ is abelian. It looks so simple as a monoid only needs to be associative and must ...
0
votes
0answers
85 views

Suppose $x^2=1$ for all $x \in G$. Prove that $G$ is abelian. [duplicate]

Suppose that $G$ is a group s.t. $x^2=1$ for all $x \in G$. Prove that $G$ is abelian. If we assume that $x^2 = 1$ for all $ x \in G$, and suppose that $a,b \in G$ and if $x=ab$ we see that $x^2 = ...
0
votes
1answer
50 views

Prove that a group $G$ is abelian [duplicate]

Suppose we have a group $(G, *).$ Prove that the group is abelian if $b * a^2 = b$ where $(a, b)$ are part of the group.
0
votes
3answers
46 views

Proving a group where every element's square is identity is Abelian [duplicate]

$G$ is a group with the property that the square of every element is the identity, then $G$ is abelian. Is my proof correct? Attempt: For every $a \in G, \space a^2 = e$ where $e$ is the identity ...

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