$D_{12}$ is isomorphic to $\Bbb Z_3\times D_4.$

Question

Prove/Disprove:

$D_{12}$ is isomorphic to $\Bbb Z_3\oplus D_4.$

$D_{12}$ has 11 elements of order $12$ whereas $\Bbb Z_3\oplus D_4$ has 9 elements of order $12$ since $\Bbb Z_3$ has $3$ elements of order $3$ and $D_4$ has $3$ elements of order $4$.

So they are not isomorphic.

Is this true?

Answer 1

I think you want to check whether $D_{12}$ is isomorphic to $\mathbb{Z}_3 \oplus D_4$ or not. The answer is no. Explanation:

Each group has $4$ elements of order $12$, $2$ elements of order $4$ and $2$ elements of order $3$. But $D_{12}$ has only $2$ elements of order $6$ whereas $\mathbb{Z}_3 \oplus D_4$ has $10$ elements of order $6$.

Further, $D_{12}$ has $13$ elements of order $2$ while $\mathbb{Z}_3 \oplus D_4$ has $5$ elements of order $2$.

You can verify this by counting number of rotations & reflections in $D_{12}$, and by taking l.c.m of orders in $\mathbb{Z}_3 \oplus D_4$.

Answer 2

I think you want to check whether $D_{12}$ is isomorphic to $\mathbb{Z}_3 \oplus D_4$ or not. The answer is no. Explanation:

Each group has $4$ elements of order $12$, $2$ elements of order $4$ and $2$ elements of order $3$. But $D_{12}$ has only $2$ elements of order $6$ whereas $\mathbb{Z}_3 \oplus D_4$ has $10$ elements of order $6$.

Further, $D_{12}$ has $13$ elements of order $2$ while $\mathbb{Z}_3 \oplus D_4$ has $5$ elements of order $2$.

You can verify this by counting the number of rotations & reflections in $D_{12}$, and by taking l.c.m of orders in $\mathbb{Z}_3 \oplus D_4$.

$D_{12}$ has a unique cyclic subgroup of order $12$ which is generated by rotation. Every cyclic group of order $n$ has $\phi(n)$ generators. Here $n =12$ and $\phi(12)=4$. To better understand $D_n$, try $D_3$ or $D_4$ or both.