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Show that a 5 holed torus is not homeomorphic to a 7 holed one

I know the fundamental groups of each of them but I don't know how to prove that they can't be isomorphic

lioness99a
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Hasan Alhajj
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  • What are their fundamental groups? Add this information to your question. – lhf Mar 13 '20 at 09:36
  • the fundemental group of 5 holed torus is 10 copies of free products of Z quotiened by the [a,b][c,d]... commutators (where first Z is generated by a second by b and so on...) and that of a seven holed we get the same but 14 copies instead of 10 – Hasan Alhajj Mar 13 '20 at 09:39
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    Please edit your question to add in any details - comments can be deleted without warning – lioness99a Mar 13 '20 at 09:45
  • Well formatted questions are better received here - a wall of text is harder to read – lioness99a Mar 13 '20 at 09:48
  • You could also try using a coarser invariant whose values are easier to compare, namely the Euler Characteristic – William Mar 13 '20 at 13:51

2 Answers2

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Indeed $\pi_1(T\# \cdots \# T)\cong \mathbb Z *\cdots *\mathbb Z/([a_1, b_1]\cdots[a_n, b_n])$. Now you can take their abelenization and apply classification of finitely generated abelian groups theorem to see that they are not isomorphic.

Semyon Abramyan
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These two surfaces are closed orientable 2-manifolds, so have genus and Euler characteristic as topological invariants. (And in this case, either distinguishes these two surfaces.)

As you observe, their fundamental groups are free abelian groups of rank $k$, where $k$ is the number of holes. You should have a theorem that states free abelian groups of differing ranks are not isomorphic.

Their homology groups are finitely generated abelian groups (i.e. finitely generated modules over a PID), so the classification theorem for finitely generated abelian groups might help. (For instance, the two groups have different ranks.)

The 7-holed torus is homoemorphic to the connected sum of a 5-holed torus and a 2-hold torus, neither of which is the 2-sphere (the identity for connected sum).

The pants decompositions of these two surfaces have distinct numbers of pairs of pants. (This is really a restatement that both are closed and their genera are different.)

Their complements in the 3-sphere are distinguishable. (Of course, each is homoemorphic to its complements in the 3-sphere, so this fact relies on some other method of discriminating these two surfaces.)

The cohomology groups of the $\require{enclose}\enclose{horizontalstrike}{n}$-holed torus, $\enclose{horizontalstrike}{T^2(n)}$ are $\enclose{horizontalstrike}{H^k(T^2(n); \Bbb{Z}) = \Bbb{Z}^{\binom{n}{k}}}$, so their first cohomology groups distinguish them.

Update: 21 August 2022

As observed by Michael Albanese, I conflated notation from whatever sources I used when this was written. For a much better discussion of these (distinct) cohomologies, see Computing the cohomology ring of the orientable genus $g$ surface by considering a quotient map. .

Eric Towers
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