A) neither \[{{H}_{1}}\cup {{H}_{2}}\]nor \[{{H}_{1}}\cap {{H}_{2}}\]is a sub group
B) nothing can be said about \[{{H}_{1}}\cup {{H}_{2}}\]and \[{{H}_{1}}\cap {{H}_{2}}\]
C) \[{{H}_{1}}\cap {{H}_{2}}\] is a sub group
D) \[{{H}_{1}}\cup {{H}_{2}}\]is a sub group
Correct Answer: C
Solution :
\[{{\text{H}}_{\text{1}}}\]and\[{{\text{H}}_{2}}=\]two subgroups of G. We know that if\[\forall a,b\in {{H}_{1}},\,ab\,\in {{H}_{1}}\] and \[\forall a,b\in {{H}_{2}},\,ab\,\in {{H}_{2}}\]and \[ab\in \,{{H}_{1}}\cap {{H}_{2}}.\] Therefore \[{{H}_{1}},\cap {{H}_{2}}\]is a subgroup.
You need to login to perform this action.
You will be redirected in
3 sec
