# Direct Product and Semi-direct Product between $S_{3}$ and $Z_{2}$, and Related Problem

I am trying to figure out the following question:

0) Is the direct product $S_{3}\times Z_{2}$ isomorphic or non-isomorphic to the semi-direct product $S_{3}\rtimes_{\phi}Z_{2}$ where $\phi:Z_{2}\rightarrow \operatorname{Aut}(S_{3}),0\mapsto Id,1\mapsto conjugation ~by~(12)$.

The only technique I can think of, beside using brutal force to write down multiplication table, is to calculate the center of these two groups:

First, it is obvious that the center of the direct product $S_{3}\times Z_{2}$ is ${(1)}\times Z_{2}$ which is of cardinality $2$.

Second, consider the center $Z$ of the semi-direct product $S_{3}\rtimes_{\phi}Z_{2}$ where $\phi$ is given as above.

The first claim I made is that: if $(a,0)\in Z$, then $a=(1)$. This can be proved by starting of assuming $(a,0)(b,0)=(ab,0)=(ba,0)=(b,0)(a,0)$ for all $b\in S_{3}$. Since the center of $S_{3}$ is trivial, we conclude that $a=(1)$.

The second claim I made is that: if $(a,1)\in Z$, then $a=(12)$. This can be proved by considering the equation $(a,1)(b,0)=(a(12)b(12),1)=(b,0)(a,1)=(b(12)a(12),1)$. Set $b=(1)$ we have $a=(12)a(12)$. It follows that $a=(1)$ or $(12)$. But if $a=(1)$, setting $b=(23)$ yields $(12)(23)(12)=(23)$ which is a contradiction. We conclude $a=(12)$ by showing that $((12),1)$ does commute with other elements in $S_{3}\rtimes_{\phi} Z_{2}$.

Now the centers of both group $S_{3}\times Z_{2}$ and $S_{3}\rtimes_{\phi}Z_{2}$ have same cardinality. I cannot decide whether they are isomorphic or not.

Any hints, ideas, thoughts are welcome.

Based on this question, I am also wondering:

1) In general how can see a semi-direct product is isomorphic to a direct product? The only result I am quite familiar with is the Proposition 11 of Section 5.5 in Dummit & Foote which says, for a semi-direct product, the following holds:

$$Id:H\rtimes_{\phi}K\rightarrow H\times K~is~isomorphism\Leftrightarrow \phi:K\rightarrow \operatorname{Aut}(H)~is~trival\Leftrightarrow K\unlhd H\rtimes_{\phi}K$$

2) Do we have any results related the question: $$(H_{1}\rtimes_{\phi_{1}}H_{2})\rtimes_{\phi_{2}}H_{3}\cong^{?}H_{1}\rtimes_{\phi_{3}}(H_{2}\rtimes_{\phi_{4}}H_{3}).$$

• Hint: 1 acts on $a \in S_3$ as $(12)a(12) = b \in S_3$. If $a = b$ for all $a$, then $a$ commutes with $(12)$ (for which there are very few candidates in $S_3$) and the map is the same as the identity map. If $a \neq b$ for some $a$, then the map is not the same as the identity map... – Eric Towers May 8 '16 at 4:12
• @EricTowers What kind of map are you talking about? – user188634 May 8 '16 at 4:26
• There's only one map in your problem and only one map in my comment: the map of $1$ acting on elements of $S_3$. (The action of $0$ is automatically trivial.) – Eric Towers May 8 '16 at 5:51
• Thank you@EricTowers. I actually made an argument to show these two groups are isomorphic. Of course I am not sure I did it correctly. May you look it below? – user188634 May 8 '16 at 6:53

Determining isomorphism classes of groups of the same order is difficult and largely an ad-hoc procedure. Looking for computable invariants is what you have to look for, such as centers, abelianizations and existence of elements of certain orders. Exactly which invariant is the "right" one isn't always easy to figure out, and takes a bit of experience and guesswork.

For direct products your (1) suffices for the "obvious" attempt to get a direct product, as if you have a semi-direct product expression then you already know the action $\phi$, which is easily verified to be (non)trivial. The tricky part happens when considering non-idenity maps, and even trickier if you don't already know a semidirect product expression. For this you may be interested in the direction extension theorem for finite groups (or groups with chain conditions):

If $G=H\times K$ is a finite group, and $H_0$ is a normal subgroup of $G$ such that $G/H_0\cong K$ and $H_0\cong H$, then $H_0$ is a direct factor of $G$. Equivalently, every short exact sequence $0\longrightarrow H\longrightarrow G\longrightarrow K\longrightarrow 0$ splits as a direct product.

You can in principle use this to verify that $G$ is not a direct product by exhibiting a case where the SES doesn't split (a semi-direct product "half-splits"). In your specific case, it suffice to find (or prove non-existence of) a group homomorphism $f\colon S_3\rtimes \mathbb Z_2\to S_3$ such that, if $g\colon S_3\to S_3\rtimes\mathbb Z_2$ is the obvious embedding, then $fg=\operatorname{id}_{S_3}$. I'll show how to find such an $f$ at the end of the answer. Normally there are many inequivalent SES's for a given $G,H,K$, but the theorem says in the direct product case they are all equivalent.

For part (2), see this MO Q&A and the comments therein. A counterexample to the general case is:

Pick $G$ dihedral of order 8, $A$ a Klein subgroup of order 4 and exponent 2, $C\subset A$ the center of $G$ (so $G/C$ is also a Klein group). Then $C$ is not part of a semidirect decomposition. – YCor

The answers consider some sufficient conditions where an iterated semidirect product may be "associative" in the sense you desire.

So consider $G=S_3\rtimes\mathbb Z_2$ with your action, and let $g\colon S_3\to G$ be the obvious injective homomorphism. As mentioned above, to show $G\cong S_3\times \mathbb Z_2$ it suffices to find $f\colon G\to S_3$ such that $f\circ g = \operatorname{id}_{S_3}$. Obviously $f$ must be surjective, so we need a normal subgroup of order 2 in $G$. The subgroups of order 2 are easily specified by the elements of order 2, which are clearly \begin{array}{ccc} ((12),0) & ((13),0) & ((23),0)\\ ((12),1) & ((13),1) & ((23),1). \end{array} The elements in the first row do not yield a normal subgroup since $S_3$ does not have normal subgroups of order $2$. The bottom left entry in the second row, however, does produce a normal subgroup. Indeed, it gives the center. The quotient thereof has image generated by the images of $((123),0)$ and $((12),0)$, which is clearly non-abelian and yields the desired $f$ in the obvious fashion. Explicitly,

\begin{align*} f((123),0) &= (123)\\f((12),0) &= (12)\\f((12),1)) &= () \end{align*} completely defines $f$. Alternatively, from the splitting condition we can also assert that the first two identities must hold for such an $f$ to exist, and then check a remaining generator (such as $((),1)$ or $((12),1)$) to see if there's a way to get a well-defined homomorphism. In this case, the action of the semi-direct product means we need the image of $((),1)$ to act by conjugation by $(12)$ in $S_3$, which will obviously happen if $((),1)\mapsto (12)$; equivalently, $((12),1)\mapsto ()$.

$\newcommand{\Span}{\left\langle #1 \right\rangle}$Let us consider the second group $H = S_{3}\rtimes_{\phi}Z_{2}$, and let us fix some notation:

• $S_{3} = \Span{a, b}$, where $a = (123)$ and $b = (12)$, and
• let $g$, an element of order $2$, be a generator of the $Z_{2}$.

By construction you have in $H$ the following conjugacy relations $$\tag{rels} a^{g} = a^{b} = a^{-1}, \qquad b^{g} = b.$$ Now consider $x = g b \notin S_{3}$. Since by (rels) $g b = b g$, we have $x^{2} = (g b)^{2} = g^{2} b^{2} = e$, the identity of the group. Also $$a^{x} = a^{g b} = (a^{g})^{b} = (a^{-1})^{b} = (a^{-1})^{-1} = a, \qquad b^{x} = b^{g b} = b,$$ that is, $x$ commutes with $a$ and $b$, and thus with all elements of $S_{3}$.

Therefore $H$ is the internal direct product $S_{3} \times \Span{x}$, which shows it is isomorphic to the first group.

After consulting my friends Weibo Fu and Angus Chung, as well as reading Eric Towers and zibadawa timmys's comment and answer above, I think I have a solution to show the two groups from 0) is isomorphic to each other. The argument proceeds as follows using inverse engineeing:

Suppose that $f:S_{3}\rtimes_{\phi} Z_{2}\rightarrow S_{3}\times Z_{2}$ is an isomorphism, we ask what properties will $f$ have?

By original post we see that the center of $S_{3}\rtimes_{\phi}Z_{2}$ is $\{((1),0),~((12),1)\}$, and the center of $S_{3}\times Z_{2}$ is $\{((1),0),~((1),1)\}$. It must be the case that:

(A) $f((12),1)=((1),1)$

The next observation is that $S_{3}\times Z_{2}$ has exactly one proper non-abelian normal subgroup, namely $S_{3}\times\{0\}$. Also, $S_{3}\times\{0\}$ is a proper non-abelian normal subgroup of $S_{3}\rtimes_{\phi}Z_{2}$. It must be the case that $f$ maps $S_{3}\times\{0\}$ onto $S_{3}\times\{0\}$.

The restriction $f|_{S_{3}\times\{0\}}:S_{3}\times\{0\}\rightarrow S_{3}\times\{0\}$ will be a group isomorphism. We know that all automorphisms of $S_{3}$ are inner automorphisms. Hence there exists some $a\in S_{3}$ such that

(B) $f(b,0)=(aba^{-1},0)$ for all $b\in S_{3}$

In $S_{3}\rtimes_{\phi}Z_{2}$, we have $((12),1)*(b,1)=((12)(12)b(12),0)=(b(12),0)=(b,0)*((12),0)$. Apply $f$ to both side, we have $$f(((12),1)*(b,1))=f((b,0)*((12),0))$$ Use the fact that $f$ is homomorphism and (A), (B): $$f((12),1)f(b,1)=((1),1)f(b,1)=f(b,0)f((12),0)=(aba^{-1}a(12)a^{-1},0)=(ab(12)a^{-1},0)$$ We get:

(C) $f(b,1)=(ab(12)a^{-1},1)$ for all $b\in S_{3}$

Now we show that:

(D) Fix arbitrary $a\in S_{3}$, $f(b,1)=(ab(12)a^{-1},1),~f(d,0)=(ada^{-1},0)$ for all $b,d\in S_{3}$ defines an isomorphism. The injectivity (hence the bijectivity) is clear. It leaves to show $f$ preserves the multiplication property. There are four cases:

Case 1: $$f((b,0)(d,0))=f(bd,0)=(abda^{-1},0)$$ $$f(b,0)f(d,0)=(aba^{-1},0)(ada^{-1},0)=(abda^{-1},0)$$

Case 2: $$f((b,1)(d,0))=f(b(12)d(12),1)=(ab(12)d(12)(12)a^{-1},1)=(ab(12)da^{-1},1)$$ $$f(b,1)f(d,0)=(ab(12)a^{-1},1)(ada^{-1},0)=(ab(12)da^{-1},1)$$

Case 3: $$f((b,0)(d,1))=f(bd,1)=(abd(12)a^{-1},1)$$ $$f(b,0)f(d,1)=(aba^{-1},0)(ad(12)a^{-1},1)=(abd(12)a^{-1},1)$$

Case 4: $$f((b,1)(d,1))=f(b(12)d(12),0)=(ab(12)d(12)a^{-1},0)$$ $$f(b,1)f(d,1)=(ab(12)a^{-1},1)(ad(12)a^{-1},1)=(ab(12)d(12)a^{-1},0)$$

We conclude $f$ is an isomorphism.

• If $S_3\times \mathbb Z_2$ has a unique proper non-abelian subgroup, consider the obvious copy of $S_3$ as well as $(123)\rtimes \mathbb Z_2\cong S_3$ in $S_3\rtimes \mathbb Z_2$. What is the subgroup of $S_3\times \mathbb Z_2$ generated by $((123),0)$ and $((12),1)$? – zibadawa timmy May 8 '16 at 7:18
• I forget to type that this subgroup is also normal. – user188634 May 8 '16 at 7:19