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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 $C_4$?
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Feb
7
revised Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
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Feb
7
revised Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
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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.
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
The thing is that it's not obvious to me that anywhere I've used the implication that $a^2$ is derivable from $a^4$. BTW, I realize that there's something wrong with my argument, but this realization relies on too much prior knowledge about group theory. The same is true of most of your answer. I'm trying to think through this problem as if I did not know any other group theory beyond what the book has presented so far (i.e. very, very little). Also, next time someone (a professor maybe, or a colleague) makes a error of arithmetic in a blackboard derivation, explain to them that the...
Feb
7
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@studiosus: I know that the relation $a^4 = 1$ does not imply the relation $a^2 = 1$. Please show me is exactly where I wrote otherwise.
Feb
7
revised Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
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Feb
6
revised Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
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Feb
6
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@user1729: will do
Feb
6
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@user1729: I figured that a hyperlink to the Google Books page for the book would give more than enough additional information.
Feb
6
reviewed Approve suggested edit on Is the sequence $ \frac{1}{10^n} $ convergent?
Feb
6
comment Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
@studiosus: I don't have that confusion, nor I see anything in my post that would suggest that I do. (And, BTW, if you're going to insult a poster's intelligence, please at least be more explicit about your reasoning.)
Feb
6
revised Why isn't $\langle a ; a^2 \rangle$ (or $\langle a;a^3, a^7\rangle$) a presentation of $C_4$?
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