# Thread: Which Group is this Group Isomorphic to?

1. ## Which Group is this Group Isomorphic to?

Hello!

Using the external direct product $\displaystyle \oplus$, the group
$\displaystyle S_3 \oplus Z_2$ is isomorphic to one (and only one!) of the following groups (this question reminds me of a game show...)

a) $\displaystyle Z_{12}$
b) $\displaystyle Z_6 \oplus Z_2$
c) $\displaystyle A_4$
d) $\displaystyle D_6$

I am assuming these groups are the well known groups
$\displaystyle Z_n$ are the integers modulo n under addittion
$\displaystyle A_n$ is the alternating group (all even permutations of n) under function composition
$\displaystyle D_n$ is the dihedral group of order $\displaystyle 2n$
and $\displaystyle S_n$ is the set of permutations of n

The question recommends determining which groups it is not isomorphic to and determining the answer by elimination.

I started by the question noting that $\displaystyle S_3 \oplus Z_2$ is cyclic, I believe? so it must be isomorphic to a cyclic group as well...
and also it must have the same number of elements of each order of any group it is isomorphic... but even armed with those facts I am having trouble finding that any of a), b), c) or d) are isomorphic to the given group $\displaystyle S_3 \oplus Z_2$

Any help appreciated!
Thank you!

2. Actually, $\displaystyle S_3\oplus \mathbb{Z}_2$ cannot be cyclic. It isn't even abelian! With this in mind, you can remove (a) and (b) from consideration.

To deal with (c) and (d), I would try looking at the orders at some elements in each group. Compare them with your $\displaystyle S_3\oplus \mathbb{Z}_2$.

3. Originally Posted by roninpro
Actually, $\displaystyle S_3\oplus \mathbb{Z}_2$ cannot be cyclic. It isn't even abelian! With this in mind, you can remove (a) and (b) from consideration.

To deal with (c) and (d), I would try looking at the orders at some elements in each group. Compare them with your $\displaystyle S_3\oplus \mathbb{Z}_2$.
Alternatively you can show that $\displaystyle S_3\oplus\mathbb{Z}_2\not\cong A_4$. To see this note that $\displaystyle S_3\oplus\{0\}\leqslant S_3\oplus\mathbb{Z}_2$ and $\displaystyle \left|S_3\oplus\{0\}\right|=6$ but $\displaystyle A_4$ has no subgroup of order $\displaystyle 6$.