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19 votes
Accepted

Is this a valid "easy" proof that two free groups are isomorphic if and only if their rank is the same?

Let me write out, in full, what the "only if" direction says: If there exists an isomorphism between the free group $F(A)$ and the free group $F(B)$ then $A$ and $B$ have the same ...
Lee Mosher's user avatar
  • 121k
7 votes

Is this a valid "easy" proof that two free groups are isomorphic if and only if their rank is the same?

Your proof is not valid. A valid proof must show that no map between them is an isomorphism; you've just shown that this particular map you've described is not an isomorphism. Note that there are maps ...
Sambo's user avatar
  • 6,356
6 votes
Accepted

A group isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3$

The isomorphisms are given by: $$\begin{align} \varphi: G\times (H\times K)&\to G\times H\times K,\\ (g, (h,k))&\mapsto (g,h,k);\\ \psi:G\times H\times K&\to (G\times H)\times K,\\ (g,h,k)&...
Shaun's user avatar
  • 45.2k
5 votes

Isomorphism theorem misunderstanding

The notation $A/B$ only makes sense (as a group) when $B$ is a normal subgroup of $A$. In general, we do not require that $H$ contains $N$ for the theorem holds. However, the equality $HN/N = \{hN : h ...
azif00's user avatar
  • 20.8k
4 votes

Is this a valid "easy" proof that two free groups are isomorphic if and only if their rank is the same?

I think you have proven that there can't be any isomorphism between $F(A)$ and $F(B)$ that maps elements of $A$ onto elements of $B$. It's natural to think that any isomorphism must be like this, but ...
Ben Millwood's user avatar
  • 14.2k
3 votes

Are these two groups $\mathbb{Q}[x]$ and $\mathbb{Q}$ under addition isomorphic or not?

It's easy to show the following proposition: Let $A, B$ be divisible and torsion-free abelian groups, then they are canonically $\mathbb Q$-vector spaces, and any group homomorphism $\rho:A\...
Just a user's user avatar
  • 15.5k
3 votes
Accepted

Subgroup of braid group $B_3$ isomorphic to itself

The covering-space argument is described in the following picture: At the top we have a 2-complex, with fundamental group $B_3$. It is precisely the presentation complex of the presentation $\langle ...
user1729's user avatar
  • 31.1k
3 votes

Subgroup of braid group $B_3$ isomorphic to itself

Anyway I managed to find out a solution: $N$ is generated by $xyxyx$ and $yxyxy$. This is found by somewhat "brute force" search by hand. I expect some more elegant explanation though.
atzlt's user avatar
  • 562
1 vote
Accepted

proving that $A_n/\operatorname{ker} \beta \cong \operatorname{ker} \alpha_{n - 1}.$

Since you have already thought about this a bit, I will give you the details. To do this canonically, let us start with the exact sequence $$A_{n+1}\xrightarrow{\alpha_{n+1}}A_n\xrightarrow{\alpha_n} ...
SomeCallMeTim's user avatar

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