1
$\begingroup$

Consider a group $G$, with subgroup $A$ and $B$ which have generators $\langle A_1,\ldots,A_m\rangle$ and $\langle B_1,\ldots,B_n\rangle$ respectively. How is the group $C$ with generators $\langle A_1,\ldots,A_m, B_1,\ldots,B_n\rangle$ related to $A$ and $B$? Is it simply the case that $C=A\times B$ or something different - either way please can you explain?

$\endgroup$
2
$\begingroup$

Consider the free group $G$ with generators $a$, $b$. This is a non-commutative group with elements the reduced words in $a$, $b$, $a^{-1}$ and $b^{-1}$, that is concatenations of these "letters" with no letter adjacent to its inverse. But $A=\langle a\rangle$ and $B=\langle b\rangle$ are each infinite cyclic (so Abelian groups). Then $C=\langle a,b\rangle =G$ is not isomorphic to $A\times B$.

$\endgroup$
0
$\begingroup$

If A and B don't sit in some larger group C, then you haven't defined a multiplication between elements of $A$ and $B$ yet so the question doesn't make sense in that case. In the example Lord Shark gave, there he defined a multiplication between the two groups, namely concatenation.

$\endgroup$
  • $\begingroup$ I have accounted for this by saying that $A$ and $B$ are subgroups of $G$. $\endgroup$ – Quantum spaghettification Jun 3 '18 at 7:27
  • $\begingroup$ My bad, missed that. In that case, your intuition should definitely not point to $A\times B$. A direct product is when you want to two groups $A,B$ to not interact at all. Since you have a multiplication already defined between the two, <A,B> will be some new group entirely. $\endgroup$ – Paul T Jun 3 '18 at 7:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.