Identify the false statement

Question

Identify the false statement (A) A non-empty subset H of a group G is a subgroup of G if and only if for every a, b ∈ H ⇒ a * b-1 ∈ H (B) The intersection of two subgroups of a group G is again a subgroup. (C) A group of order three is not a belian. (D) If in a group G,(ab)2 = a2b2 ∀ a, b ∈ G, then G is abelian

Tardigrade Admin · Accepted Answer

A group of order three is not abelian is not true. If O(G) = prime, then G is always abelian and O(G) ≤ 6 always abelian group.

Q. Identify the false statement

A non-empty subset $H$ of a group $G$ is a subgroup of $G$ if and only if for every $a, b \in H \Rightarrow a * b^{- 1} \in H$

The intersection of two subgroups of a group $G$ is again a subgroup.

A group of order three is not a belian.

If in a group $G, (ab)^{2} = a^{2} b^{2} \forall a, b \in G$ , then $G$ is abelian

A group of order three is not abelian is not true.
If $O (G)$ = prime, then G is always abelian and $O (G) \leq 6$ always abelian group.

Q. Identify the false statement

A non-empty subset H of a group G is a subgroup of G if and only if for every a,b∈H⇒a∗b−1∈H

The intersection of two subgroups of a group G is again a subgroup.

A group of order three is not a belian.

If in a group G,(ab)2=a2b2∀a,b∈G, then G is abelian

A group of order three is not abelian is not true. If O(G) = prime, then G is always abelian and O(G)≤6 always abelian group.

A non-empty subset $H$ of a group $G$ is a subgroup of $G$ if and only if for every $a, b \in H \Rightarrow a * b^{- 1} \in H$

The intersection of two subgroups of a group $G$ is again a subgroup.

If in a group $G, (ab)^{2} = a^{2} b^{2} \forall a, b \in G$ , then $G$ is abelian

A group of order three is not abelian is not true.
If $O (G)$ = prime, then G is always abelian and $O (G) \leq 6$ always abelian group.