# A set containing more than half elements of a group [duplicate]

I wish to prove the exercise which states that for a set $A$ containing more than half elements of a group $G$, every element of $G$ is a product of two elements of $A$.

My attempt:

By Lagrange Theorem, subgroup generated by $A$ must coincide with $G$, so every element in $G$ is a product of elements from $A$. How can I prove that the product is of exactly two elements?

## marked as duplicate by Derek Holt group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 7 '16 at 18:06

Let $g\in G$. Then $A$ and $gA^{-1}$ are two sets larger than half of $G$, hence intersect. But if $a_1=ga_2^{-1}$ then $g=a_1a_2$