I am just learning generators and relations basics. A clarification

Let $G$ be a finite group in which every element(non-zero) is a possible candidate to be in the minimal generating set $S$ of the group. (What I mean here by a possible candidate is, for e,g., in $D_5 = \langle r,s| r^5 = s^2 =e, s^{-1}rs = r^{-1}\rangle$, I can very well replace $r$ by $r^2, r^3$ etc and express $D_5$ with the same above relations). With that understanding, I call $r^2$, $r^3$ etc as candidates for being generators of $S$.

In such a set up, suppose there exists a finite group where all non-zero elements in the group are probable candidates for generators, is my following claim true?

Claim: Let the minimal cardinality of the generating set $S$ of group $G$ be $k$. Then, order of $G$ can only be $p_1p_2\dots p_k$.

My argument goes as below:

Let $S = {a_1, a_2, ..., a_k}$ be the minimal generating set to generate the group $G$. $G = \langle S\rangle$. Now, take any generator $a_i$ of $G$. Let $m_i$ be its order. My claim is that $m_i$ is a prime number. Else, then $\exists 1< d_i$ < $m_i$ that divides $m_i$. Clearly, $a_i^{(m_i/d_i)}$ $\in$ $\langle a_i\rangle$ and it generates only a proper sub group of $\langle a_i\rangle$ which is not possible as $a_i^{(m_i/d_i)}$ is also a generator of G and needs to compensate for the loss of $a_i$ from the generating set. Thus, $\langle(a_i)^{(m_i/d_i)}\rangle = \langle a_i\rangle$. This cannot happen as order of $(a_i)^{(m_i/d_i)}$ is $d_i$, which is clearly less than $m_i$. So, $m_i$ is prime $\forall$ $i$ from $1$ to $k$. Thus, $|G| = m_1m_2\dots m_k$ = $p_1p_2\dots p_k$.

Note: Please don't criticize the proof straight away without properly understanding what I am trying to achieve. Please check first if my conclusion has a possibility to be true even though my proof is not fool proof. I would welcome counter examples very happily.

P.S: is there any known result regarding the upper bound for $k$ (I mean apart from $|G| - 1$)?

  • $\begingroup$ The upper bound for $k$ is $\log _2 |G|$, and this is attained in an elementary abelian group of order $2^k$. $\endgroup$
    – Derek Holt
    Jan 3 '15 at 9:09

Given any non-zero element $g$ of $S_n$ when $n$ is not $4$ there exists another element that generates $S_n$ along with $g$. For a proof look at theorem $A$ of this article. Therefore a group with that property and $k=2$ can have any order that is a factorial (and possibly many other orders). The point is that the number of primes in the order of a factorial is unbounded.

  • $\begingroup$ So shall I conclude the following statements 1) $S_n$ is a finite group in which the only non-generator is id (so that it becomes an example of a group I was referring to, above). 2) $S_n$ can have more than 2(k=2 for $S_n$) primes as factors of its order, say I take $|S_5|$ = $2^3*3*5$. These two conclusions say my claim is wrong. $\endgroup$
    – user166305
    Jan 3 '15 at 22:28
  • $\begingroup$ Yeah, that's precisely what I was trying to say. :) $\endgroup$
    – Asinomás
    Jan 4 '15 at 17:41

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