The group $S_3 \oplus \mathbb{Z}_2$ is isomorphic to one of the following: $\mathbb{Z}_{12}, \mathbb{Z}_6\oplus\mathbb{Z}_2, A_4, D_6$. Determine which by elimination.
I have already determined that it is not isomorphic to $\mathbb{Z}_{12}$.
I am not understanding the work for $A_4$ and $D_6$ if someone could please explain step-by-step?
On
$(12)(23)=(123)$ while $(32)(21)=(321)$, and these are different permutations in $S_3$. Embedding, $S_3 \hookrightarrow S_3 \oplus \mathbb Z_2$ by $a \mapsto (a,0)$, we can see that $S_3 \oplus \mathbb Z_2$ is nonabelian, while $ \mathbb Z_6 \oplus \mathbb Z_2$ is clearly abelian, since the additive structure is given pointwise.
To see that an isomorphism perserves abelian properties, consider a map $\phi:\mathbb Z_6 \oplus \mathbb Z_2 \to S_3 \oplus \mathbb Z_2$. By surjectivity, there exists $a,b \in \mathbb Z_6 \oplus\mathbb Z_2 $ so that $\phi(a),\phi(b)=(12),(23)$. But $(12)(23)=\phi(ab)=\phi(ba)=(32)(21)$, which is a contradiction.
Elements in $A_4$ are even permutations of $4$ "letters" (here represented by the numerals $1$ through $4$, but any $4$-element set would do), that is, permutations in $S_4$ that are representable as a product of an even number of transpositions.
The cycle-types of $S_4$ (which correspond to partitions of $4$) are:
a) $4$-cycles-these require a product of three transpositions (at least), and so are not elements of $A_4$-for example: $(1\ 2\ 3\ 4) = (1\ 4)(1\ 3)(1\ 2)$.
b) Two disjoint $2$-cycles-these are elements of $A_4$: for example $(1\ 2)(3\ 4)$.
c) $3$-cycles-these are elements of $A_4$, e.g.: $(1\ 2\ 3) = (1\ 3)(1\ 2)$.
d) $2$-cycles, or transpositions. One is odd, so transpositions are not elements of $A_4$.
e) The identity, which moves no letters. Zero is an odd number, so the identity is in $A_4$.
So $A_4$ consists of cycle-types b), c) and e). Every element of $A_4$ is of one of those three forms. Now the order of a product of disjoint cycles is the least common multiple of the cycle lengths. So the order of a $3$-cycle is $3$, and the order of two disjoint transpositions is $2$. So we have no element of order $6$ in $A_4$ (the order of the identity is, of course, $1$).
However:
$((1\ 2\ 3), 1) \neq (e, 0)$ (its order is not $1$)
$((1\ 2\ 3), 1)^2 = ((1\ 3\ 2), 0) \neq (e, 0)$ (its order is not $2$)
$((1\ 2\ 3), 1)^3 = (e, 1) \neq (e, 0)$ (its order is not $3$)
$((1\ 2\ 3), 1)^4 = ((1\ 2\ 3), 0) \neq (e, 0)$ (its order is not $4$)
$((1\ 2\ 3), 1)^5 = ((1\ 3\ 2), 1) \neq (e, 0)$ (its order is not $5$)
$((1\ 2\ 3),1)^6 = (e,0)$, the order is $6$.
So $A_4$ has no element of order $6$, but $S_3 \oplus \Bbb Z_2$ does, so they cannot be isomorphic.
To settle the matter, we ought to exhibit an isomorphism. If we express $D_{12}$ as:
$D_{12} = \langle r,s: r^6 = s^2 = 1, sr = r^{-1}s\rangle$,
we need to find a suitable map $\phi: S_3 \oplus \Bbb Z_2 \to D_{12}$.
First, it should be obvious that $S_3\oplus \Bbb Z_2 = \langle ((1\ 2\ 3), 0), ((1\ 2), 0) ,(e, 1)\rangle$. So to show that:
$S_3\oplus \Bbb Z_2 = \langle S\rangle = \langle ((1\ 2\ 3), 1), ((1\ 2), 1)\rangle$ we just need to show the elements $((1\ 2\ 3), 0), (1\ 2), 0)$ and $(e, 1)$ are generated by $S$.
Two are easy:
$((1\ 2\ 3), 1)^3 = (e,1)$
$((1\ 2\ 3), 1)^4 = ((1\ 2\ 3), 0)$
The last one is just a tad harder:
$((1\ 2\ 3), 1)^3((1\ 2), 1) = (e,1)((1\ 2), 1) = ((1\ 2), 0)$
So now, if we call $a = ((1\ 2\ 3), 1)$ and $b = ((1\ 2), 1)$, we have:
$S_3\oplus \Bbb Z_2 = \langle a,b: a^6 = b^2 = (e,0)\rangle$, which looks promising, we might conjecture we can define $\phi$ by sending $a \mapsto r$ and $b \mapsto s$, and extend by $\phi(a^kb^m) = \phi(a)^k\phi(b)^m$.
To confirm this is then a homomorphism (and so an isomorphism), it suffices to show $ba = a^{-1}b$ (why?).
We compute: $ba = ((1\ 2), 1)((1\ 2\ 3), 1) = ((2\ 3), 0)$
$a^{-1}b = ((1\ 3\ 2), 1)((1\ 2),1) = ((2\ 3), 0)$, and we are done.