Is it true that in an abelian group, if $a^{5} = b^{7} = (ab)^{11}=e$ then $a=b=e$?

Question

It seems that this should be true, but doing operations with these expressions does not appear something useful, perhaps we can use the fact that the order of the group should be divided by 35, but I do not know how to develop a thought

You say "abelian" in the title, but not in the post or the tags. The relevant information should be included in the post itself. — Arturo Magidin, Dec 14 '23 at 18:28

score 5 · Accepted Answer · answered Dec 14 '23 at 18:36

5

In an abelian group:

If $a^5=e$ and $b^{7}=e$, then note that $a^{11}$ has the same order as $a$ (since if $x$ has order $n$, then $x^k$ has order $n/\gcd(n,k)$, and the order of $a$ is either $1$ or $5$), and likewise $b^{-11}$ has the same order as $b$.

If $(ab)^{11} = a^{11}b^{11}=e$, then $a^{11}=b^{-11}$. But the order of $a^{11}$ is either $1$ or $5$; and the order of $b^{-11}$ is either $1$ or $7$; thus, both their orders must be $1$, so $a=b=e$.

Note that commutative is required. It is known that for any $r,s,t\gt 1$, there exists a group that contains elements $a$ and $b$ such that the order of $a$ is $r$, the order of $b$ is $s$, and the order of $ab$ is $t$.

answered Dec 14 '23 at 18:36

Arturo Magidin

417,286

2

Why $a^{11}$ "has the same order as" $a$, when we know that $a^{11}=a$? – Anne Bauval Dec 14 '23 at 18:47
@AnneBauval I don't understand your question. If $a^{11}=a$, then don't they have the same order? Sure, you can conclude it differently, but is there an error in the justification I give, in the parentheses right after the assertion? That way the argument is the same for $a$ and for $b$. – Arturo Magidin Dec 14 '23 at 18:48
Nono, no error! just a suggestion of stylistic simplification. – Anne Bauval Dec 14 '23 at 18:50
@AnneBauval Right; but since I will use the $\gcd$ argument for $b$, might as well use it for both. – Arturo Magidin Dec 14 '23 at 18:50
I agree, it just looked strange at first sight. – Anne Bauval Dec 14 '23 at 18:51

Shaun · Answer 2 · 2023-12-14T20:25:46.210

2

Assuming the group is abelian, $(ab)^{11}=e$ implies $a^{11}=b^{-11}$, so $a=b^3$. But then

$$\begin{align} e&=a^{5}\\ &=(b^{3})^5\\ &=b^{15}\\ &=(b^7)^2b\\ &=b. \end{align}$$

So yes.

edited Dec 14 '23 at 20:25

answered Dec 14 '23 at 18:36

Shaun

47,747

score 1 · Answer 3 · answered Dec 14 '23 at 19:49

It might be worth noting that this kind of question can be answered at least two ways as corollaries of the structure theorem for finitely-generated modules over a PID. (In particular, for abelian groups, such questions are definitely answerable, unlike more general "word problems" in non-abelian groups.) Let's write things additively, rather than multiplicatively, and reframe the question as asking whether $(5,0)$, $(0,7)$, and $(11,11)$ generate all of $\mathbb Z^2$.

An algorithmic approach uses the fact that $\mathbb Z$ is Euclidean, so that the following approach will terminate: doing a sort of row reduction to the matrix $\pmatrix{5&0\cr 0&7\cr 11&11}$, first subtract $2$ times the top row from the third, then $5$ times the third from the first, etc. Eventually, we obtain $\pmatrix{0&0\cr 0&1\cr 1&0}$ (or equivalent), so those three vectors do generate the whole $\mathbb Z^2$.

More directly (but not so algorithmically), consider the $3$ two-by-two minors, $35,55,77$. Their gcd is $1$, so (by the theorem, and a little exterior algebra) those three "vectors" generate all of $\mathbb Z^2$.

Is it true that in an abelian group, if $a^{5} = b^{7} = (ab)^{11}=e$ then $a=b=e$?

3 Answers3