I have to prove that if $G$ and $G'$ are two finites group of same cardinal, then they are isomorphic.
Actually, it looks obvious. Suppose $G=\{g_1,...,g_n\}$ and $G'=\{h_1,...,h_n\}$. Does the homomorphism $g_i\longmapsto h_i$ work ?
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2This is not true in general., For example $\Bbb Z/4\Bbb Z$ is not isomorphic to $\Bbb Z/2\Bbb Z\times \Bbb Z/2\Bbb Z$. – Gregory Grant Feb 10 '16 at 20:51
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There must be another assumption - I'm guessing both groups are cyclic? – Gregory Grant Feb 10 '16 at 20:51
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For cyclic, it's definitely true. Map generator to generator. – Sinister Cutlass Feb 10 '16 at 20:52
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Is it possible you meant "sets" instead of "groups"? As everyone is pointing out, the statement is far from true for groups. – lulu Feb 10 '16 at 20:55
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I am pretty sure there are at least two non-isomorphic groups of any cardinality except for $0,1$ and $p$ (where $p$ is prime). – Asinomás Feb 10 '16 at 21:01
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1Also, why do you think $g_i \to h_i$ is a homomorphism? – Feb 10 '16 at 21:01
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@dREaM: Overly fancy by a lot, but one can use Upward Lowenheim-Skolem. – André Nicolas Feb 10 '16 at 21:06
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@dREaM : there is only 1 group of order 15, isn't it? – Watson Feb 10 '16 at 21:07
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Oh, you seem to be right. – Asinomás Feb 10 '16 at 21:12
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I think we also need $pq$ where $p<q$ and $q$ is not congruent to $1\bmod p$. – Asinomás Feb 10 '16 at 21:17
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@dREaM : it is possible to prove that the number of groups of order $n$ (up to isomorphism) is $1$ iff $(n,\phi(n))=1$. Therefore, I think the condition should be $(n,\phi(n))>1$. – Watson Feb 10 '16 at 21:30
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Why are people downvoting this person? They asked a legitimate question and showed their attempted answer. So downvoting without constructive criticism is rude in my opinion. – Gregory Grant Feb 10 '16 at 21:52
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This is not true in general., For example $\Bbb Z/4\Bbb Z$ is not isomorphic to $\Bbb Z/2\Bbb Z\times \Bbb Z/2\Bbb Z$.
You probably also need to assume both groups are cyclic. In that case you can write one as $\{a,a^2,\dots,a^n\}$ and the other as $\{b,b^2,\dots,b^n\}$. Then map $a\mapsto b$ to get the isomorphism.
Gregory Grant
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Ok, in fact I had to say if this assertion is wrong or false. And in fact it false for the reason you said. Thank you – MSE Feb 10 '16 at 21:01
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No, it doesn't work since this result is wrong. For example $\mathbb Z/4\mathbb Z$ is not isomorphic to $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$. To prove that there are not isomorphic, simply remark that $\mathbb Z/4\mathbb Z$ has an element of order $4$ whereas $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$ has no element of order $4$.
Surb
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