Help resolving this contradiction in descriptions of the fundamental groups of the figure eight and n-torus

I'm faced with a few contradictory statements:

1) This article on Wolfram MathWorld says that "$\mathbb{Z}*\mathbb{Z}=\lbrace(n,m)\rbrace$ is a free abelian group of rank 2." (I thought that this is the definition of $\mathbb{Z}\times\mathbb{Z}$.)

2) This article also on Wolfram MathWorld says that the fundamental group $\pi_1$ of the figure eight is "$\mathbb{Z}*\mathbb{Z}$" and that the first homology group $H_1$ of the figure eight is "$\mathbb{Z}\times\mathbb{Z}$", implying that these two groups are different, i.e., not isomorphic. This article also states the fundamental and first homology groups of a n-torus $\mathbb{T}^n$ are both "$\mathbb{Z}^n$" (which I thought usually refers to $\mathbb{Z}\times . . . \times\mathbb{Z}$ n times?).

3) This article also on Wolfram MathWorld states that the fundamental group of the figure 8 is the "free group with two generators" (i.e., is NOT the free abelian group of rank 2).

4) My textbook, Munkres, states in Section 60 that the fundamental group of the figure eight is not abelian and that the fundamental group of the two-holed torus (2-torus or $\mathbb{T}^2$) is not abelian.

5) This answer on math.stackexchange.com says that the fundamental group of the n-torus is $\langle{a_1}, b_1, ..., a_n, b_n \mid [a_1, b_1]...[a_n, b_n]=0 \rangle$, where $[a, b]=aba^{-1}b^{-1}$.

6) Section 60 of Munkres also states that the fundamental group of a 1-torus is $\mathbb{Z}\times\mathbb{Z}$, which is what I consider the free abelian group of rank 2 (correct me if I'm wrong). As a sanity check, I tested whether this is isomorphic to the group that results from statement (5) when $n=1$, and found that it is.

The best way to resolve this contradiction seems to be to say that statement (1) was a typo and should have said the free abelian group of rank 2 is $\mathbb{Z}\times\mathbb{Z}$, NOT $\mathbb{Z}*\mathbb{Z}$. To fully resolve the contradiction I would then guess that $\mathbb{Z}^n$ in statement (2) is also a typo (since thisusually refers to $\mathbb{Z}\times . . . \times\mathbb{Z}$ n times), and should instead be $\langle{a_1}, b_1, ..., a_n, b_n \mid [a_1, b_1]...[a_n, b_n]=0 \rangle$, as in statement 5. Interpreting things this way, I would also guess that the notation $\mathbb{Z}* . . . *\mathbb{Z}$ refers to the free group with n generators (can someone at least verify whether this is true or not?).

Does this resolution of the contradiction (which implies there are two flagrant typos in Wolfram MathWorld articles) seem correct? Would really appreciate any help interpreting these statements and correcting my understanding, as well as any corrections to things I've said that are wrong!

• The first article on MathWorld uses a very unusual notation for the free abelian group. As you noticed, this object is usually denoted $\mathbb{Z}\times\mathbb{Z}$ in the case of rank $2$. This is not the same as the free (in general non-abelian) group $\mathbb{Z}\ast\mathbb{Z}$. – asdq Jan 14 '18 at 20:54
• Regarding 2) the article is right. The fundamental group of the $n$-torus is indeed $\mathbb{Z}^n$ (meaning the direct product) since $\mathbb{T}^n=(S^1)^n$ and the fundamental group functor commutes with direct products. – asdq Jan 14 '18 at 20:57
• @asdq, but $\mathbb{T}^1=S^1\times{S}^1$, which is not $(S^1)^1$, right? – Rasputin Jan 14 '18 at 21:02
• @Rasputin: no, it's the thing asdq said. – Qiaochu Yuan Jan 14 '18 at 21:02
• Right, this makes sense now, you all are right that I was confusing the "n-torus" with the "n-holed torus"/"genus n torus" – Rasputin Jan 14 '18 at 21:09

In general, MathWorld has all sorts of mistakes in it, and unlike Wikipedia, they can't be corrected by people who find them. It's really not a trustworthy source.

1) This notation is highly nonstandard. $\mathbb{Z} \ast \mathbb{Z}$ usually denotes the free product of $\mathbb{Z}$ with itself. This is the free group $F_2$ on two generators, and in particular is nonabelian.

2) All of this is correct. In particular, the fundamental group of the $n$-torus really is $\mathbb{Z}^n$.

3) This is correct.

4) This is correct. "Two-holed torus" does not refer to $\mathbb{T}^2$, it refers to the surface of genus two.

5) This refers to the surface of genus $n$, not to $\mathbb{T}^n$.

6) This is correct.

So, there were several obstacles for you here: not just MathWorld's confusing notation, but also the distinction between the "$n$-torus" and the "$n$-holed torus" (personally I think this is confusing terminology for exactly this reason, and so I prefer to use "surface of genus $n$" instead).

• Thanks for this response! Makes a lot more sense now, the terminology for the "n-torus" and "n-holed torus" is definitely a bit confusing haha. – Rasputin Jan 14 '18 at 21:11
• Not "orientable surface of genus $n$"? – Eric Towers Jan 14 '18 at 23:41
• @Eric: I prefer not to use genus for non-orientable surfaces at all, although that would be a less ambiguous term. – Qiaochu Yuan Jan 14 '18 at 23:42
• When you say 2) is all correct, are you disregarding the definition of $\mathbb Z*\mathbb Z$ from 1)? – stewbasic Jan 15 '18 at 2:19
• @stewbasic: yes, it's a bad definition and I'm using the standard one. – Qiaochu Yuan Jan 15 '18 at 5:28

Don't believe everything you read on Wolfram. Normally $\Bbb Z\ast\Bbb Z$ denotes the free product (or coproduct in the category of groups) of $\Bbb Z$ and $\Bbb Z$. This is the free group on two generators and is non-Abelian. The fundamental group of the figure-eight is $\Bbb Z\ast\Bbb Z$ and is non-Abelian. Its first homology group is the Abelianisation of $\Bbb Z\ast\Bbb Z$, which is $\Bbb Z\times\Bbb Z$.

The fundamental group of the $n$-torus (which is $(S^1)^n$) is $\Bbb Z^n$. This isn't to be confused with the orientable surface of genus $n$, (sometimes called an "$n$-holed torus") whose fundamental group is non-Abelian for $n\ge 2$ and which has the presentation given in (5).

In short, Wolfram's MathWorld has typos and more serious errors too. Unlike in Wikipedia, users cannot correct them.

• Thanks for the response! Your response and Qiaochu's really cleared this up. Too bad there is no easy way to correct the errors in Wolfram. – Rasputin Jan 14 '18 at 21:13