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I am trying to teach myself group theory and I recently came across the topic of Isomorphisms. I know that 2 groups are isomorphic if there is a one-on-one correspondence between their elements. So if the groups have a different order, does that mean they are not isomorphic? Such as a group $S$ and its permutation group $S_{n}$ for $n>1$.

José Carlos Santos
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  • Possibly helpful: https://math.stackexchange.com/questions/2039702/what-is-an-homomorphism-isomorphism-saying/2039715#2039715 – Ethan Bolker Nov 06 '17 at 18:55
  • You said "one-on-one" which I've never heard before, but also the terminology seems confusing here. According to Wikipedia "one-to-one" means injective but "one-to-one correspondence" is bijective. This may be the source of your confusion. Better to use injective/surjective terminology IMO. – Era Nov 06 '17 at 19:56
  • @Era In the US (at least where I study), the terms "one to one" and "onto" are much more common than "injective" and "surjective". "One to one correspondence" is definitely much more confusing than "bijective", though, so we tend to use the latter. – Logan Clark Nov 14 '17 at 14:43

2 Answers2

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The answer is “yes”, but your definition of isomorphism is not correct. An isomorphism between groups is a bijection $\varphi$ which preserves the product, that is, such that $\varphi(x.y)=\varphi(x).\varphi(y)$ for every $x$ and every $y$ in the domain. But since it must be a bijection then, yes, groups with distinct orders cannot be isomorphic.

José Carlos Santos
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If two groups are isomorphic then their orders are the same. It follows from the fact that isomorphism between groups is a bijection.