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visits member for 3 years, 5 months
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I'd rather walk than take a taxi.


1d
awarded  Explainer
Sep
28
awarded  Revival
Sep
25
comment Is it true to say that “it's not logically possible to prove something can't be done”?
I am not sure that I agree with your interpretation of the strip (although really I don't know). If your interpretation were correct then would Dan not be saying "and I can prove it using a computer"? (Although computer's are Turing machines...perhaps human minds are just Turing machines? Thus my earlier disclaimer: really I don't know...)
Sep
24
awarded  Autobiographer
Sep
24
comment Equation in Free Group
Your reason should be "...there do not exists non-trivial universal equalities valid in any finite group with a fixed number of generators", no?
Sep
24
comment Group extensions, Geometric group theory
Sorry, yes, I was thinking of 2-complexes. Yes, the 3-cells do affect homology - if I remember correctly, this is how Eilenberg–MacLane spaces are constructed (by attaching the appropriate $n$-cells).
Sep
23
comment Group extensions, Geometric group theory
I don't have the time to look into this fully, but by "attaching appropriate 3-cells", do they not just mean filling in the relators? Isn't that just the presentation complex? (I haven't actually read your link - I am just ad-libbing...)
Sep
19
comment Exercise on generated subgroup
What have you tried? If you tell us this then we will be better able to help you. And it helps us feel that we are not just doing your homework for you.
Sep
18
comment Doubt about isomorphic groups!
I had never even contemplated the possibility that $K$ would not be normal! +1, just for pointing that out!
Sep
18
comment To which group is this presentation isomorphic?
What have you tried? If you tell us this then we will be better able to help you. And it helps us feel that we are not just doing your homework for you.
Sep
18
revised Do all Groups have a representation?
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Sep
18
revised Do all Groups have a representation?
added 62 characters in body
Sep
18
revised Do all Groups have a representation?
added 62 characters in body
Sep
17
awarded  Disciplined
Sep
9
revised How is $A \implies B$ different from $B \implies A$
edited title
Sep
9
revised If $G$ is a finite group whose $p$-Sylow subgroup $P$ lies in its center, then there is a normal subgroup $N$ of $G$ with $P\cap N=\{e\}$ and $PN=G$
edited title
Sep
9
comment When can a homomorphism be determined entirely by its generators
@WantTobeAbstract Your map is for all group elements, not just the generators! If it was for just the generators, then you first need a presentation of your group $G$. The relations are the relations of this presentation. For example, If $G$ was the group $\mathbb{Z}_6\times\mathbb{Z}_2=\langle a, b; ab=ba, a^6=1, b^2=1\rangle$ and $n=2$ then you need to verify that $a^2b^2=b^2a^2$, that $a^{12}=1$ and that $b^4=1$. These three relations all clearly hold, so the map $a\mapsto a^2$, $b\mapsto b^2$ is a homomorphism.
Sep
8
comment When can a homomorphism be determined entirely by its generators
@WantTobeAbstract Ah, okay, you just need to check that the homomorphism works for the relators of a given presentation (with your given generators). You might find this answer of mine useful.
Sep
8
comment When can a homomorphism be determined entirely by its generators
I am unsure precisely what you are asking, but I think the restriction you are after is if your map is $\phi: G\rightarrow H$ then you want the generators to generate a subgroup of $H$ which is isomorphic to a homomorphism of $G$. In my first example, $\mathbb{Z}$ contains no elements of finite order but all homomorphic images of $\mathbb{Z}_n$ are finite cyclic. In the third example, $G$ has only itself and the trivial group as a homomorphic image, but $H$ does not contain a copy of $G$.
Sep
8
revised When can a homomorphism be determined entirely by its generators
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