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If we have two groups $G,G^{'} $ of order $n$ and both have the same number of elements of a given order then are the two groups isomorphic?

I meant that if $o(G)=o(G^{'})=n$ and if $G$ has $p$ elements of order $m$ then $G^{'}$ has also $p$ elements of order $m$ and this holds for each $m\in \Bbb N$ , then are $G,G^{'} $ isomorphic?

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    You mean $G$ has $p$ elements of order $m$ if and only if $G'$ has $p$ elements of order $m$ for every $m$? – 2017-01-14
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    They are **finite** groups, aren't they? – 2017-01-14
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    Yes they are@ajotatxe – 2017-01-14
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    Seems like induction on order of the groups and quotients sholud do the work... (Only guessing). – 2017-01-14
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    Strongly related: https://math.stackexchange.com/questions/729611, https://math.stackexchange.com/questions/1296833 – 2017-01-14
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    In the Abelian case, yes, in the non-Abelian case, no. See http://math.stackexchange.com/questions/693163/groups-with-same-number-of-elements-of-each-order and http://math.stackexchange.com/questions/72944/two-finite-abelian-groups-with-the-same-number-of-elements-of-any-order-are-isom – 2017-01-14

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