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Show that $D_{12}$ is isomorphic to $D_6 \times C_2$, where $D_{2n}$ is the dihedral group of order $2n$ and $C_2$ is the cyclic group of order $2$.

I'm somewhat in over my head with my first year groups course. This is a question from an example sheet which I think if someone answered for me could illuminate a few things about isomorphisms to me.

In this situation, does one use some lemma (that the direct product of two subgroups being isomorphic to their supergroup(?) if certain conditions are satisfied)? Does $D_{12}$ have to be abelian for this? Do we just go right ahead and search for a fitting bijection? Can we show the isomorphism is there without doing any of the above?

If someone could please answer the problem in the title and talk their way through, they would be being very helpful.

Thank You.

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5  
What are $D_{12}$ and $D_6$? –  Eric O. Korman Jan 12 '12 at 0:27
2  
They can be the same, due to the lack of consensus about how to name the dihedral groups. Some name them for the number of elements in the group; others count the number of corners in the regular polygon they are the symmetry groups of. So $D_{12}$ and $D_6$ are both possible names for the symmetry group of a regular hexagon. (But it sounds fairly unlikely that an exercise would shift notation in mid-stride like this). –  Henning Makholm Jan 12 '12 at 0:32
    
Sorry, I completely messed the question up - I forgot to write in an essential part. I plead mental fatigue. I've corrected the question. I need to show D_12 isomorphic to D_6xC_2 –  user27182 Jan 12 '12 at 0:34

4 Answers 4

up vote 5 down vote accepted

You've gotten several good answers. Let me address the questions in your third paragraph.

I believe the result you are looking for is the following:

Definition. Let $G$ be a group, and let $N$ and $M$ be normal subgroups of $G$. We say that $G$ is the internal direct product of $N$ and $M$ if and only if:

  1. $N\cap M = \{e\}$; and
  2. $G=NM$.

And then we have:

Theorem. Let $G$, $H$, and $K$ be groups. The following are equivalent:

  1. $G$ is isomorphic to the external direct product of $H$ and $K$, $G\cong H\times K$.
  2. There exist normal subgroups $N$ and $M$ of $G$ such that $N\cong H$, $M\cong K$, and $G$ is the internal direct product of $N$ and $M$.

(There is no requirement that $G$ be abelian).

Proof. To show that (1) implies (2), let $\varphi\colon H\times K \to G$ be the isomorphism. Then let $N=\varphi(H\times\{e\})$, $M=\varphi(\{e\}\times K)$; since $\varphi$ is an isomorphism, and $H\times K = (H\times\{e\})(\{e\}\times K)$ and both subgroups are normal, condition (2) follows.

Conversely, let $\phi\colon N\to H$ and $\psi\colon M\to K$ be isomorphisms, and assume that $G$ is the internal direct product of $N$ and $M$. Define a map $f\colon G\to H\times K$ as follows: given any $g\in G$, then since $G=NM$ we have $g=nm$ for some $n\in N$ and $m\in M$; define $f(g) = (\phi(n),\psi(m))$.

First, note that this is well-defined: if $nm=n'm'$ for $n,n'\in N$ and $m,m'\in M$, then $(n')^{-1}n = m'm^{-1}$; but this element is in both $N$ and $M$, and since $N\cap M=\{e\}$, it follows that $m'm^{-1}=(n')^{-1}n = e$, hence $m=m'$ and $n=n'$. Thus, the expression $g=nm$ is unique, and since $\phi(n)\in H$ and $\psi(m)\in K$, we have that $f(g)$ is well defined and lies in $H\times K$.

Next, note that if $n,\in N$ and $m\in M$, then the fact that $N\cap M=\{e\}$ implies that $nm=mn$: because $$n^{-1}m^{-1}nm = \left( n^{-1}m^{-1}n\right)m = n^{-1}\left(m^{-1}nm\right),$$ the middle expression is in $M$, the third expression in $M$; so $n^{-1}m^{-1}nm\in N\cap M = \{e\}$, hence $n^{-1}m^{-1}nm = e$, so multiplying on the left by $mn$ we get $nm=mn$, as claimed.

Thus, to show $f$ is a group homomorphism, let $g,g'\in G$; write $g=nm$ and $g'=n'm'$. Then $gg' = (nm)(n'm') = n(mn')m' = n(n'm)m' = (nn')(mm')$. Hence, $$f(gg') = (\phi(nn'),\psi(mm')) = (\phi(n)\phi(n'),\psi(m)\psi(m')) = (\phi(n),\psi(m))(\phi(n'),\psi(m')) = f(g)f(g').$$

Next, to show $f$ is one-to-one, note that if $f(g)=(e,e)$, where $g=nm$, then $\phi(n)=e$ so $n=e$ (since $\phi$ is an isomorphism), and $\phi(m)=e$ so $m=e$. Thus, $g=nm=ee=e$. So $\mathrm{ker}(f)=\{e\}$, hence $f$ is one-to-one.

Finally, to show that $f$ is onto, let $h\in H$ and $k\in K$. Since $\phi$ and $\psi$ are isomorphisms, there exist $n\in N$ and $m\in M$ such that $\phi(n)=h$ and $\psi(m)=k$. Then $f(nm) = (\phi(n),\psi(m)) = (h,k)$. So $f$ is onto. Thus, $f$ is an isomorphism, and we are done. $\Box$

So, in order to show that $D_{12} \cong D_6\times C_2$, you could find normal subgroups $N$ and $M$ of $D_{12}$ such that $N\cong D_6$, $M\cong C_2$, $N\cap M=\{e\}$, and $D_{12}=NM$.

You don't tell us how you think about $D_{12}$; one can think of it as permutations on the set $\{1,2,3,4,5,6\}$ (number the vertices of the regular hexagon and describe the elements of $D_{12}$ by how the vertices are permuted), or abstractly as a group presented as: $$D_{12} = \Bigl\langle r,s\;\Bigm|\; r^6 = s^2 = 1,\quad sr = r^{-1}s\Bigr\rangle$$ (or maybe other ways of presenting $D_{12}$; thought I think that is the most common one).

Consider the subgroup $N = \langle r^2,s\rangle$. It is not hard to verify that this subgroup has $6$ elements, and so, being of index $2$, it must be normal in $D_{12}$. Then let $M=\{1,r^3\}$. Since $r^3r = rr^3$ and $r^3s = sr^3$, then $M$ is central, and hence normal (verify that $x^{-1}r^3x = r^3$ for every $x\in D_{12}$ using the observations above). And so $M$ is normal and isomorphic to $C_2$.

Then show that $N\cong D_6$ (which is not hard).

Finally, note that $N\cap M = \{e\}$ and therefore $|NM| = |N||M| = 12 = |G|$; since $N$ and $M$ are normal, $NM$ is a subgroup, and so $NM=G$. Thus, you will get, using the theorem, that $G$ is isomorphic to $D_6\times C_2$.

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Too complicated for OP's question, but VERY interesting. (EDIT : just saw that he actually asked for this.)The fact that you've shown something useful to any group theorist is cool. +1! Great answer. –  Patrick Da Silva Jan 12 '12 at 13:58
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@PatrickDaSilva: Part of the complication arises from the fact that I in-lined the proof that if $N$ and $M$ are normal and $N\cap M = \{e\}$, then $nm=mn$ for all $n\in N$ and $m\in M$. –  Arturo Magidin Jan 12 '12 at 17:06
    
It was necessary, that's what I edited my comment. I was saying that it was too complicated for an answer because of its generality, but when I saw that was OP's request, I just kept the cool part of my comment. =) –  Patrick Da Silva Jan 12 '12 at 22:03
    
This is a really useful answer, thank you. In fact, the theorem you presented is one from my notes which I thought I should be using for this question - only in my notes there is a confused mistake which says G must be abelian. Does G = NM mean every element of G is of the form nm with N in N and m in M? –  user27182 Jan 14 '12 at 23:28
    
@Bryn: It's possible that the result in your notes is in a different context; you don't actually need $G$ to be abelian, but of course it could be that $G$ is abelian (in which case, you would not need to check normality). As to your second question: yes; given a group $G$, and subsets (not necessarily subgroups) $A$ and $B$ of $G$, $AB$ is the set $\{ab\mid a\in A,b\in B\}$ of all products of an element of $A$ with an element of $B$. If $A$ and $B$ are subgroups, then it is a nice exercise to show that $AB$ is a subgroup if and only if $AB=BA$ as sets; if either is normal, this will happen. –  Arturo Magidin Jan 14 '12 at 23:58

It's very hard to know exactly how you are supposed to attack such a problem without knowing your background. For example, if you've classified all groups of order 12, then this is a simple problem of showing they both match the same group in the list.

Here's an approach from a generator and relation viewpoint.

$D_{2n}$ is generated by a rotation of $360/n$ degrees (or $2\pi/n$ radians) and any reflection. Let's call such a rotation $x$ and reflection $y$. Then $x^n=1$ (I'll call the identity "$1$") and $y^2=1$ (because $(360/n)\times n = 360 = 0$ degrees and any reflection is its own inverse). Then because any rotation times any reflection is a reflection, we have $xy$ is a reflection and thus $xyxy=1$. This implies (keeping in mind $y=y^{-1}$ since $y^2=1$) that $xy=yx^{-1}$ and $yx=x^{-1}y$. Thus given any string of $x$'s and $y$'s we can push all of the $x$'s to the left and all of the $y$'s to the right. So an arbitrary element looks like $x^iy^j$. Keeping in mind that $x^n=1$ and $y^2=1$ so that $x$'s exponents "work mod $n$" and $y$'s "work mod $2$", we have $$D_{2n} = \langle x,y \;|\; x^n=y^2=xyxy=1 \rangle = \{ 1,x,\dots,x^{n-1},y,xy,x^2y,\dots,x^{n-1}y \}$$ (all $2n$ elements accounted for).

In particular, $D_{12} = \langle x,y \;|\; x^6=y^2=xyxy=1 \rangle =\{1,x,\dots,x^5,y,xy,\dots,x^5y\}$ and (using different letters to avoid confusion) $D_{6}= \langle a,b \;|\; a^3=b^2=abab=1 \rangle = \{1,a,a^2,b,ab,a^2b\}$. Also, let $C_2 = \langle u \;|\; u^2=1 \rangle = \{1,u \}$.

To establish an isomorphism you'll need to come up with a bijection that respects these relations. Let's try to invent a map $\varphi:D_{12} \to D_6 \times C_2$.

Notice that $x$ has order 6. So it needs to map to an element of order 6. Now $a$ has order 3 (the largest order of any element in $D_6$) and $u$ has order 2. So $(a,u) \in D_6 \times C_2$ has order $|(a,u)|=\mathrm{lcm}(3,2)=6$. Thus we can map $\varphi(x)=(a,u)$. It then seems logical to send $y$ (an element of order 2) to $b$ (an element of order 2) paired with the identity: $\varphi(y)=(b,1)$.

So if $\varphi$ is to be a homomorphism, we must have: $\varphi(x^iy^j)=\varphi(x)^i\varphi(y)^j=(a,u)^i(b,1)^j=(a^i,u^i)(b^j,1)=(a^ib^j,u^i)$.

Ok. Now we've invented our candidate map: $\varphi(x^iy^j)=(a^ib^j,u^i)$. Let's prove that it's an isomorphism.

  • Homomorphism: $\varphi(x^iy^j\cdot x^ky^\ell) = \varphi(x^ix^{-k}y^jy^\ell)=\varphi(x^{i-k}y^{j+\ell})=(a^{i-k}b^{j+\ell},u^{i-k})$. On the other hand, $\varphi(x^iy^j)\varphi(x^ky^\ell)=(x^iy^j,u^i)(x^ky^\ell,u^k)=(x^iy^jx^ky^\ell,u^iu^k)=(x^ix^{-k}y^jy^\ell,u^{i+k})$ $=(x^{i-k}y^{j+\ell},u^{i+k})$. Finally, notice that $(i+k)-(i-k)=i-i+k+k=2k$ so that $i+k=i-k$ mod $2$. Thus $u^{i-k}=u^{i+k}$. Therefore, $\varphi(x^iy^j\cdot x^ky^\ell) = \varphi(x^iy^j)\varphi(x^ky^\ell)$. So we've shown $\varphi$ is a homomorphism.

  • One-to-One and Onto can be seen many different ways (one way is just to map all 12 elements and see that you do indeed have a bijection). Another: Onto because $x$ maps to $(a,u)$ and $y$ maps to $(b,1)$ and these two elements: $(a,u)$ and $(b,1)$ generate the codomain. One-to-one: Suppose $\varphi(x^iy^j)=(1,1)$ then $(a^ib^j,u^i)=(1,1)$. Thus $u^i=1$ so that $i$ is even. $a^ib^j=1$ implies that $i$ is divisible by 3 and $j$ is even. But if $j$ is even, then $y^j=1$ and $i$ divisible by 3 and 2 means it's divisible by 6 so $x^i=1$. Thus $x^iy^j=1$. Thus $\varphi$ has a trivial kernel so it's one-to-one.

Well, there you go. It's definitely not the most efficient route, but hopefully this will give you some stuff to digest.

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Have you thought about an isomorphism for these two? By the way, $D_6$ is only abelian when $n = 2$ (It gives you a group of order $4$ in which $rs = sr^{-1} = sr$.), so forget the $D_{12}$ abelian idea.

Try to exploit properties of isomorphisms. The center of the group is preserved by isomorphisms, and the center of $D_{12}$ is $\{ 1, r^3 \}$, hence you must map $r^3$ to the element that generates the center in the other group ; that is, $(1, x)$ (that is if I choose to write $C_2 = \langle x \ | \ x^2 = 1 \rangle$). Now the element $s$ in $D_{12}$ has the property that $ys = sy^{-1}$ for every element $y \in D_{12}$, and the element $(s,0)$ in $D_6 \times C_2$ also has this property, so map $s$ to $(s,0)$. What about $r$ in $D_{12}$? It is tempting to map it to a power of $r$ in $D_6$, but which one? Make things simple and try mapping it to $r$. Now we have a "guess" : \begin{align} r & \mapsto (r,1) \\ r^3 & \mapsto (1,x) \\ s & \mapsto (s,1) \\ \end{align}

and extend by multipliying those together

(ex : $r^5 s = r^3 r^2 s \mapsto (1,x)(r^2,1)(s,1) = (r^2s,x)$.)

Showing that this is an isomorphism is now calculations. I leave that up to you. Just so you feel that I am right, for instance $$ (r^4 s)(rs) = r^4 r^{-1} s^2 = r^3 \mapsto (1,x) = (rssr^{-1},x) = (rsrs,x) = (rs,x)(rs,1). $$ Another nice way to guess this is to use geometry ; looking at the symmetries of the hexagonal plate as $D_{12}$ and looking at $D_6 \times C_2$ as the symmetries of the triangular plate inscribed in the hexagonal plate, in which $C_2$ gives you the half-turn symmetry of the hexagonal plate.

Hope that helps,

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+1 Nice answer. Sometimes for me the best way is to construct directly the isomorphism to remove all doubts. –  user38268 Jan 12 '12 at 8:23

Assuming $D_{n}$ is the dihedral group of order $n$, I would proceed as follows. Note that $D_{6} \cong S_{3}$, and $S_{3}$ is generated by $(12)$ and $(123)$. Therefore $D_{6} \times C_{2} = \langle ((12),[0]),((123),[1]) \rangle$. Next note that $D_{12} = \langle r,s | \, r^{6}=s^{2}=e , s^{-1}rs=r^{-1} \rangle$, map the generators of $D_{6} \times C_{2}$ to the generators of $D_{12}$, and show this extends to a bijection.

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You wrote $D_{12}$ as $D_{24}$. –  Patrick Da Silva Jan 12 '12 at 3:29
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Whoops, I did, and it looks like the Wikipedia page about the dihedral group needs some editing too. –  Jackson Walters Jan 12 '12 at 3:34
    
It is not standard to say $D_n$ or $D_{2n}$. But wrote $D_6$ as the dihedral group with $6$ elements and $D_{12}$ as the dihedral group with $24$ elements, so that must generate some confusion. –  Patrick Da Silva Jan 12 '12 at 3:48

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