# If two finite groups have the same number of elements of the same order, does this necessarily imply they are isomorphic? [duplicate]

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I am trying to prove a rather difficult question and I have arrived at a small proof that I can prove true after thoroughly (and exhaustively) analyzing the group structure.

My question is, if two finite groups have the same number of elements of the same order... are they necessarily isomorphic? If not, what are some properties of the group structure that can show that two groups are necessarily isomorphic?

## marked as duplicate by miracle173, Dietrich Burde 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(); } ); }); }); Apr 25 '17 at 9:18

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## 1 Answer

In general, finite groups of the same order are not isomorphic. Indeed, this is far from true. So there are, for example, 49 487 365 422 pairwise non-isomorphic groups of order $1024$. This is a lot.

On the other hand, two cyclic groups of the same order are indeed isomorphic.

• The OP accepted your answer but actually this is not an answer to his question. Because he wants "the same number of elements of the same order" – miracle173 Apr 25 '17 at 6:27
• @miracle173 It was the answer to the original question. I have also answered the new question at this duplicate. – Dietrich Burde Apr 25 '17 at 8:14
• Thank you, I think it is the duplicate of the question you linked to. – miracle173 Apr 25 '17 at 9:12