MHB Proof of G Abelian When 3-abelian & 5-abelian

  • Thread starter Thread starter mathbalarka
  • Start date Start date
  • Tags Tags
    Proof
AI Thread Summary
If a group G is both 3-abelian and 5-abelian, it can be proven that G is abelian. The proof begins by demonstrating that square elements in G commute, leading to the conclusion that all elements of G commute with each other. This is established through various manipulations of the group operations, showing that powers of elements maintain commutativity. Ultimately, the properties of 3-abelian and 5-abelian groups imply that any two elements a and b in G satisfy the relation ab = ba. Therefore, G is confirmed to be an abelian group.
mathbalarka
Messages
452
Reaction score
0
A group G is both 3-abelian and 5-abelian, then prove that G abelian in general.

Balarka
.
 
Mathematics news on Phys.org
Here is my solution :

$(ab)^3 = a^3\cdot b^3$ implies $(ab)^2 = b^2\cdot a^2$ and $(ab)^5 = a^5\cdot b^5$ implies $(ab)^4 = b^4\cdot a^4$.

Using both of these, we conclude that the square elements in $G$ commutes, i.e, $a^2\cdot b^2 = b^2\cdot a^2$.

Now, $(ab)^3 = (ab)\cdot(ab)^2 = (ab)\cdot(b^2\cdot a^2) = (ab)\cdot(a^2\cdot b^2) = a^3\cdot b^3 = a\cdot(a^2 * b^2)\cdot b$, after elimination by right and left cancellation laws, we reckon that the square elements of $G$ commutes with any other elements of $G$, i.e. $a^2\cdot b = b\cdot a^2$.

Since $G$ is 3-abelian, every 3rd power and 2nd power commutes, i.e, $a^3\cdot b^2 = a^2\cdot b^3$. Since $G$ is also 5-abelian, $a^5\cdot b^4 = b^4\cdot a^5$.

It can also be derived that $a^4\cdot b^3 = b^3\cdot a^4$ by the previous property that the square element commutes with any other element of $G$.

Now $a^4\cdot b^3 = b^3\cdot a^4$ implies $b\cdot(a^4\cdot b^3) = b^4\cdot a^4 = a^4\cdot b^4$.

Hence, by associativity, $(ba)\cdot(a^3\cdot b^3) = a^4\cdot b^4 = (ba)^4$, so, $a^3\cdot b^3 = (ba)^3$.

Note that by definition of $G$, $(ba)^3 = b^3\cdot a^3$, hence the cube elements in $G$ commute too.

Consider $b^4\cdot a^5 = a^5\cdot b^4 = a^2\cdot(a^3\cdot b^3)\cdot b = a^2\cdot(b^3\cdot a^3)\cdot b = (a^2\cdot b^3)\cdot(a^3\cdot b) = (b^3\cdot a^2)\cdot(a^3\cdot b) = b^3\cdot(a^5\cdot b)$.

By left cancellation law, $b\cdot a^5 = a^5\cdot b$. From this, we can easily derive that $b\cdot a^3 = a^3\cdot b$.

If we write this explicitly, $b\cdot a^3 = b\cdot (a^2\cdot a) = (b\cdot a^2)\cdot a = (a^2\cdot b)\cdot a = a^2 (b\cdot a) = a^3\cdot b$.

By left cancellation, $b\cdot a = a\cdot b$, hence $G$ is commutative.
 
Last edited by a moderator:
so $(ab)^5 = a^5b^5 $ implies $(ba)^4 = a^4b^4$ (1)
and $(ab)^3 = a^3b^3 $ implies $(ba)^2 = a^2b^2$ (2)

so $baba*(ba)^2 = baba*aabb$
using (1) $baba*aabb = a^4b^4$, so
using (2), $aabb*aabb = a^4b^4$or
$bbaa = a^2b^2$
plugging this into (2)
$baba = bbaa$
or ab = ba.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Back
Top