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?
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!
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 ...
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.
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.
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 $ab=babaab$ $=$...
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 ...
205 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.
$G$ 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 ...