The following argument works if G is finite. In this case since f is a monomorfism is a biyection.

Since f is a morphism we have:

(ab)^3=f(ab)=f(a)f(b)=a^3b^3

from this;

(ab)^2=a^2b^2 (0)

On the oher hand:

(aya^{-1})^3=f(a)f(y)f(a)^{-1}=a^3y^3a^{-3} (1)

but also:

(aya^{-1})^3=ay^3a^{-1} (2)

from (1) and (2) we have: a^2y^3=y^3a^2 (3)

Since f is a bijection b^2=y^3 for some y so from (3) a^2b^2=b^2a^2

Finally replacing in (0):

(ab)^2=a^2b^2=b^2a^2

abab=baba

ba=ab

Nov 30, 2012

Ostap Chervak · Ivan Franko National University of Lviv

From (ab)^3 = ababab = aaabbb we get that
baba = aabb, so (ba)^4 = (aabb)(aabb)=(a^2 b^2) (a^2 b^2) = (a^2 b^2)^2 = b^4 a^4=
bbbbaaaa=bb(abab)aa=b (ba)^3 a
So babababa=bbababaa.
Now we multiply this equality by b^-1 on the left and a^-1 on the right, we get
ababab=bababa => (ab)^3 = (ba)^3 hence by injectivity of f we obtain ab=ba so group is commutative.

Note that you must require that f in injective: Take group of nine elements {1, x, y, xx, yy, xy, xxy, xxyy, xyy} with multiplication defined by yx = xyy.

## All Answers (2)

Luis Fuentes García· University of A CoruñaSince f is a morphism we have:

(ab)^3=f(ab)=f(a)f(b)=a^3b^3

from this;

(ab)^2=a^2b^2 (0)

On the oher hand:

(aya^{-1})^3=f(a)f(y)f(a)^{-1}=a^3y^3a^{-3} (1)

but also:

(aya^{-1})^3=ay^3a^{-1} (2)

from (1) and (2) we have: a^2y^3=y^3a^2 (3)

Since f is a bijection b^2=y^3 for some y so from (3) a^2b^2=b^2a^2

Finally replacing in (0):

(ab)^2=a^2b^2=b^2a^2

abab=baba

ba=ab

Ostap Chervak· Ivan Franko National University of Lvivbaba = aabb, so (ba)^4 = (aabb)(aabb)=(a^2 b^2) (a^2 b^2) = (a^2 b^2)^2 = b^4 a^4=

bbbbaaaa=bb(abab)aa=b (ba)^3 a

So babababa=bbababaa.

Now we multiply this equality by b^-1 on the left and a^-1 on the right, we get

ababab=bababa => (ab)^3 = (ba)^3 hence by injectivity of f we obtain ab=ba so group is commutative.

Note that you must require that f in injective: Take group of nine elements {1, x, y, xx, yy, xy, xxy, xxyy, xyy} with multiplication defined by yx = xyy.

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