# Let $G$ be Abelian. Then any subgroup of $G$ is normal. Does the converse hold? [duplicate]

I need a little help with the following problem of abstract algebra:

Let $$G$$ an Abelian group. Clearly, any subgroup of $$G$$ is normal. Is the opposite true, that is if every subgroup of $$G$$ is normal, then $$G$$ is Abelian?

## marked as duplicate by Morgan Rodgers, José Carlos Santos, rschwieb, MathematicsStudent1122, Derek Holt abstract-algebra 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(); } ); }); }); Mar 6 at 17:03

• Can you add some of your own thoughts to how to proceed? Where are you getting stuck? – gt6989b Mar 6 at 16:49
• Have you tried looking at some small nonabelian groups to find a counterexample? Or tried to start a proof? – Morgan Rodgers Mar 6 at 16:49
• I see,can you tell me some examples of non-abelian group to check please? – José Marín Mar 6 at 16:57
• Please do not title your questions with useless things like "Help with Abstract Algebra". In a case like this, the actual question you are addressing is a good title. – rschwieb Mar 6 at 17:01

The converse is not true. Think of the quaternion group of order $$8$$, $$Q=\{\pm 1,\pm i,\pm j,\pm k\}$$. Can you see why this gives a counterexample?
• No. Let $H$ be a subgroup of $G$. Then for sure we must have $|H|\le |G|$. That is the reason we have "sub" here. $H$ must be smaller than $G$ in some sense. Also Lagrange proved that $|H|$ must divides $|G|.$ – Tortuga Mar 6 at 17:53