Prove that the groups $D_6$ and $\mathbb{Z}_6$ are isomorphic

Question

I'm stuck trying to prove that:

Prove that the groups $D_6$ and $\mathbb{Z}_6$ are isomorphic.

My attempt: Let $D_6=\{e,a,a^2,b,ab,a^2b\}$ where $a^3=b^2=e$ and $ba=a^2b$ and $\mathbb{Z}_6=\{[0],[1],[2],[3],[4],[5]\}$

Let $f:\mathbb{Z}_6 \rightarrow D_6$ such that

$[0] \rightarrow e$
$[1] \rightarrow a$
$[2] \rightarrow a^2$
$[3] \rightarrow b$
$[4] \rightarrow ab$
$[5] \rightarrow a^2b$

Constructed in this way it is clear that f is bijective.

I'm stuck trying to prove that $f(a\cdot b) = f(a)*f(b)$ where $(\cdot)$ is the modular sum in $\mathbb{Z}_6$ and $*$ is the product in the dihedral group can someone help me?

Answer 1

I have no wonder you can't solve this problem, because your statement is wrong.

If $\Bbb Z_6$ and $D_6$ are isomorphic, then $D_6$ must be abelian. But $(ba)(ab)^{-1}=a^2b(ab)^{-1}=a$, so $ab\ne ba$, a contradiction.

Answer 2

In $\Bbb Z_6$ we have the element $[1]$ which is of order 6. Yet the maximal order of an element in $D_6$ is 3: for example the order of $a$ the 60 degree rotation.

Since isomorphism preserves the order of elements (nice exercise) we can deduce that both groups aren't isomorphism to each other.

Fun fact: this is a special case of a $pq$ group. For two primes $p<q$ such that $p\mid q-1$, there are exactly 2 groups of order $p \cdot q$ (up to an isomorphism), one of which is Abelian while the other isn't. This fact can be proved using the Sylow theorems.