4,222 reputation
41339
bio website
location
age
visits member for 3 years, 6 months
seen Jan 3 at 16:19

Mar
7
comment Natural uses for the co-product of sets?
@MJD: I think that tagging may be an inessential "implementation detail", an artifact of adopting a "synthetic", as opposed to an "analytic", point of view; I've written at length about this in an "Epilogue" to my original question.
Mar
7
revised Natural uses for the co-product of sets?
added 4744 characters in body
Mar
7
revised Natural uses for the co-product of sets?
added 4744 characters in body
Mar
3
accepted Natural uses for the co-product of sets?
Mar
3
revised Looking for proof of monicity of co-product insertions
added 2 characters in body; edited title
Mar
3
asked Natural uses for the co-product of sets?
Mar
3
reviewed Approve The following subset is a subspace of $\mathbb{R}^2$
Mar
3
reviewed Reject Intuition behind failure rate.
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@anon: I see your point. Also Callus just pointed out an even bigger hole in my reasoning. I think that I can straighten out the mess in my head now. Thanks again.
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@Callus: OK! I see! That's it. Thank you!
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@anon: the book (whether correctly or not) writes "...use $b^8$ and $b^{12}$ as defining relators, and show that $b^4$ can be derived. Hence, by the preceding exercise [i.e. exercise 4], $b^8$ and $b^{12}$ are a set of defining relators." Are you saying the book is wrong, or are you saying that I am misinterpreting it? (I figure it's the latter, but if so I have to concede that I just don't get it, and that it's completely beyond my ken.) Anyway, thanks for your patience. (And likewise to Callus too.)
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@Callus: You may be right. I find the text very vague on the subject of picking relators. For one thing, it discusses the defining of relators in terms mappings from sets of symbols to group elements. If one uses the mapping $a \mapsto 2 \in \mathbb{Z}_4$, then $a^2 = 1$ (i.e. $2 + 2 = 0$ in $\mathbb{Z}_4$), so $a^2$ is a relator of the cyclic group of order $4$. And I can derive $a^4$ from it.
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@anon: the "then by exercise 4 blah blah" bit comes basically verbatim from the book. You can see the exact quotation in my post.
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@anon: the reasoning is exactly the same one that the authors give in the hint to exercise $5$(c). In a nutshell: 1) pick some relators $x, y, z...$; 2) show that $b^4$ can be derived from $x, y, z, ...$; 3) then, by Exercise 4, $x, y, z...$ are defining relators of $C_4$. Now, for my $x, y, z...$ I pick the single relator $b^2$, proceed to derive $b^4$ from it, and conclude that $\langle b; b^2 \rangle$ (i.e. $\langle a; a^2 \rangle$) is a presentation of $C_4$.
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@Callus: you're right; I really wish I could delete the post, and maybe do it over...
Feb
7
revised Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
deleted 380 characters in body
Feb
7
revised Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
added 89 characters in body
Feb
7
revised Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
edited title
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@Callus: Thanks, I think my error is along the lines of what you write in your comment, but when I re-read the bit in the book that gave rise to this question, I remain confused. I give more details in the latest additions to my post.
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
...reason is that they are "confused about addition", or subtraction, or whatever the operation was where the misstep happened, and see if they don't take your "explanation" as an insult.