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Let $G$ be a finite group and $A=\{a_{1},...a_{k}\}$ and $B=\{b_{1},...,b_{k}\}$ be two minimal generating sets of $G$ such that $|a_{i}| = |b_{i}|$ for $i=1,\dots,k$. We define $\alpha(a_{i})=b_{i}$. Is it an automorphism?

(At first I thank Derek Holt and Tobias Kildetoft for their explanations. Now I changed the conditon to minimal generating sets.

If $a_{1}^{t_{1}}...a_{k}^{t_{k}}=1 $ for $t_{i}<\mid a_{i}\mid$, then $t_{1}=t_{2}=...t_{k}=0$. Also we have this condition for the set B.)

Now I think with this condition added $\alpha$ is automorphism. Thank you

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The same example as in your previous question makes this fail. You would need some sort of independence of the generators to make it work even for abelian groups. –  Tobias Kildetoft Feb 7 at 12:42
    
Another illustrative example: If the group is elementary abelian, then having this property would imply that the group of automorphisms acts $(|G|-1)$-transitively on the non-identity elements, which is only the case very rarely (one can determine precisely when) –  Tobias Kildetoft Feb 7 at 12:46
    
In the current setup, you can take the set of all elements of the group as your $a_i$ and $b_i$, suitable permuted. To avoid this, you can require that the generating set is minimal (though I am not sure minimality will be sufficient for this except in special cases). –  Tobias Kildetoft Feb 7 at 12:48
    
You can generate $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/6\mathbb{Z}$ either by an element of order 2 and one of order 6, or by two elements of order 6. –  Alex B. Feb 7 at 13:08
    
Take $\mathbb Z_2\times\mathbb Z_3$ with minimal generating set $\{(1,0), (0,1)\}$. Swapping those two generators can't give you an automorphism because they are of different order, so how could this work in the abelian case in general? –  Christoph Feb 7 at 13:09

3 Answers 3

up vote 3 down vote accepted

Here is a generally true version: every group has a presentation where one lists generators and certain relations that they have amongst each other. Given any two generating subsets of the group that satisfy those defining relations, there is an automorphism that takes one set to the other.

For finite abelian groups one can always take the presentation to be a minimal generating set with the relations only the order relations and the obvious commutativity relations (which are automatic on any subset of the abelian group). Compare to Derek Holt's example: $C_2 \times C_4 \times C_4 = \langle x \rangle \times \langle y \rangle \times \langle z \rangle$ has presentation $$\left\langle x,y,z ~\middle|~ x^2 = y^4 = z^4 = 1, yx=xy, zx=xz, zy=yz \right\rangle$$ So absolutely any generating set $(a,b,c)$ with orders 2,4,4 will define an automorphism via $x\mapsto a, y\mapsto b, z \mapsto c$.

However, not all minimal generating sets satisfy these relations. For instance $a=xy,b=y,c=z$ and $a'=xz, b'=y, c'=z$ are both minimal generating sets with orders 4,4,4 but $a \mapsto a', b \mapsto b', c\mapsto c'$ runs into trouble as $x = ab^3 \mapsto a' (b')^3 = xy^3z$, but the first has order 2 and the second has order 4, an impossibility for a homomorphism. These generating sets don't satisfy the defining relations, so we get into trouble. The first set does satisfy these relations, but notice the extra bold one!

$$C_2 \times C_4 \times C_4 = \left\langle a,b,c ~\middle|~ a^4 = b^4 = c^4 = 1, \mathbf{a^2=b^2}, ba=ab, ca=ac,cb=bc \right\rangle$$

The generating set $(a',b',c')$ does not satisfy these relations (well not the bold one at least), so the automorphism does not exist.

This generator and relations method works for arbitrary groups as well. For instance the simple group of order 60 has a nice presentation $$\left\langle a,b ~\middle|~ a^2 = b^3 = (ab)^5 \right\rangle$$ where all the relations are orders, but not necessarily orders of generators. At any rate, any two distinct elements $a\neq b$ of any group that satisfy these relations generate a subgroup isomorphic to $A_5$, and given any two pairs $a\neq b$ and $a'\neq b'$ that both satisfy the relations, the function $a\mapsto a', b \mapsto b'$ defines an isomorphism between $\langle a,b\rangle$ and $\langle a',b'\rangle$.

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It's still not true even for abelian groups. Let $G = C_2 \times C_4 \times C_4 = \langle x \rangle \times \langle y \rangle \times \langle z \rangle$. Then $\{xy,y,z\}$ and $\{ xz,y,z\}$ are both minimal generating sets, but there is no automorphism $\alpha:xy \mapsto xz, y \mapsto y, z \mapsto z$, because it would have to map $x$ to $xy^{-1}z$, which has order $4$.

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Here is a case where it does hold:

If $G$ is elementary abelian, then the generating sets correspond to bases over a suitable prime field (due to minimality), and the statement now just becomes the familiar argument from linear algebra that one can send any basis to any other basis via an invertible linear map.

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