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- Mar 22, 2013

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A group G is both 3-abelian and 5-abelian, then prove that G abelian in general.

Balarka

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Balarka

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- Thread starter mathbalarka
- Start date

- Thread starter
- #1

- Mar 22, 2013

- 573

A group G is both 3-abelian and 5-abelian, then prove that G abelian in general.

Balarka

.

Balarka

.

- Thread starter
- #2

- Mar 22, 2013

- 573

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.

$(ab)^3 = a^3\cdot b^3$ 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:

- Jan 11, 2013

- 68

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.