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Suppose I have a group $G$, given as finitely many generators and finitely many relations, say. Furthermore, suppose I have two finite sets $A,B$ of elements of the group. I would like to know if the subgroups generated by $A$ and $B$ are the same, and more generally if the normal subgroups generated by $A$ and $B$ (set of all elements formed by products of generators and their conjugates) are the same.

Perhaps I can find a way of writing elements of $A$ in terms of elements of $B$ (and their conjugates) and vice versa, in which case I'm done. Alternatively, if I can come up with a homomorphism to some simpler group (like the abelianization) under which the images of $A$ and $B$ clearly generate different groups, I have my answer. But both of these rely on my ingenuity; is there a more systematic way of working it out?


An example (sorry I can't come up with something simpler which isn't obvious)!: $$G=\langle a_1,b_1,a_2,b_2 \, | \, [a_1,b_1][a_2,b_2]=1 \rangle \\ A = \{ [a_1,b_1]b_1 b_2^{-1}, a_1 a_2 \} \\ B= \{ [a_1,b_1] b_2^{-1} b_1, a_1 [a_1,b_1] a_2 [a_1,b_1]^{-1}\} $$ Do $A$ and $B$ generate the same normal subgroup of $G$?

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Virtually all questions involving groups defined by finite presentations are theoretically undecidable in general, so you cannot hope to find a general procedure for solving these sorts of problems.

The answer to your specific question is no. I adjoined the generators of $A$ to the relations of $G$ to get the quotient of $G$ by the normal closure of $A$, computed the $3$-quotient to class $3$, which had order $19683$, and found that the image of the element $[a_1.b_1]b_2^{-1}b_1$ for $B$ was nontrivial. (I did that in a CAS of course!)

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  • $\begingroup$ Thanks, that's what I suspected and feared! What software did you use? Even if I can't do it in general it would be nice to have a decent tool to help me in specific cases. $\endgroup$ – Holographer Jul 26 '14 at 21:48
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    $\begingroup$ I used Magma but you would probably find it easier to use GAP, which you can just download and easily install. It can compute finite $p$-quotients to any class, which is often effective for proving negatives. $\endgroup$ – Derek Holt Jul 26 '14 at 21:50
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I'm not so sure what you mean. There seem to be a great number of questions in your post.

Here's what I'll answer:

Suppose I have a group G, given as finitely many generators and finitely many relations, say (What?). Furthermore, suppose I have two finite sets A,B of elements of the group. I would like to know if the subgroups generated by A and B are the same, and more generally if the normal subgroups generated by A and B (set of all elements formed by products of generators and their conjugates) are the same.

Let's use an example. Let your group be $\mathbb{Z}$ under addition. It is generated by $\{-1,1\}$. So now let $A = \{-3,3\}$, and let $B = \{-5,5\}$. $A$ and $B$ generate the subgroups $3\mathbb{Z}$ and $5\mathbb{Z}$, respectively. These are clearly not the same. Both $A$ and $B$ are normal in $\mathbb{Z}$.

The normal subgroups of $A$ and $B$ will clearly not be the same. (E.g. consider the subgroup of $A$ generated by $\{-6,6\}$, which is normal in $A$, and which is not the same as the subgroup of $B$ generated by $\{-10,10\}$, which is normal in $B$).

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  • $\begingroup$ I haven't done any group theory in a while, so I may be a bit rusty! I'd appreciate if someone else read over my answer and confirmed that my work is correct. $\endgroup$ – Newb Jul 26 '14 at 21:40
  • $\begingroup$ I give an example at the bottom. When the group is finitely-generated abelian, it's pretty straightforward; harder when things don't commute! $\endgroup$ – Holographer Jul 26 '14 at 21:45
  • $\begingroup$ @Holographer Oh. You should have said that you're interested in non-commutative groups. $\endgroup$ – Newb Jul 26 '14 at 21:46

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