Talk:Generating set of a group/Archive 1
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| Archive 1 |
Non-generators
Could we have an example of a non-trivial non-generator? AxelBoldt 18:31, 6 October 2002 (UTC)
- I was going to create a new entry for Frattini subgroup (Fr(G) is the intersection of all maximal subgroups of G) which would include examples, as well as a proof that Fr(G) is the set of non-generators; this was intended as a "teaser" to get people to jump to that link. A simple example would be the cyclic Cp², where p is prime; then p is a non-generator. User:chas_zzz_brown 18:52, 6 October 2002 (UTC)
Removed intro statement
I removed this statement from the intro:
- Whilst this may appear to produce excessively long strings of terms, in fact, these strings often can be reduced.
This is true, I'm sure, but quite out of place when we're not even done defining things yet. If someone could expand and reword, it could be turned into a paragraph appearing later in the article. Deco 18:56, 14 July 2005 (UTC)
Number of elements term
What is the technical term for the number of elements of a (minimal) generating set?
- I'm not aware of any. Around the web I see things like "size of the minimal [or minimum] generating set". Deco 03:03, 28 July 2005 (UTC)
- A group may have minimal generating sets of several different cardinalities (and some infinite groups have no minimal generating set at all), so "number of elements of a minimal generating set" is not well-defined. I'm guessing that what you really meant is the least cardinality for a generating set of the group; I've seen this referred to as the generating rank of the group (at least for abelian groups). --Zundark 20:23, 26 November 2005 (UTC)
- Maybe you should look at the paper: Nonorientable embeddings of groups. European J. Combin. 9 (1988) 445-461. There the authors distinguish between irredundant generating set and minimal generating set. Since they work with Cayley graphs they give involutary generators different weight when defining the rank of the group. Tomo 21:33, 26 November 2005 (UTC)
Why "finitely many"?
I don't see why the first line stipulates finitely many elements. Does this not mean that an infinite group is not generated by a set of its elements (for instance, that the integers are not generated by 1)? I would be inclined to drop this stipulation; any comments? Mcboogie 18:49, 3 October 2007 (UTC)
- I would be inclined to drop it too. It's correct (if correctly understood), but redundant (since infinite products aren't defined), and the fact that you've misunderstood it shows that it's misleading. --Zundark 19:05, 3 October 2007 (UTC)