# Proving that the groups $S_3$ and $D_6$ are isomorphic [duplicate]

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In particular, $S_3$ is the group of permutations of $\{1,2,3\}$, and $D_6$ is the dihedral group of symmetries of the triangle (written as $D_{2\cdot 3}$). In generator-relation form, $D_6 = \left< r,s \mid r^3 = s^2 = e, rs = sr^{-1} \right>$.

I can intuitively see very easily that the two groups are isomorphic; the symmetries of a triangle can be viewed as orderings of the vertices (1,2,3).

However, proving this rigorously, I need some help.

Is the following proof correct?

Let $f : D_6 \to S_3$ be a homomorphism. Define $f(r) = (123)$ and $f(s) = (12)$. Then $$f(r)f(s) = (123)(12) = (13) = f(rs)$$

and we need this to equal $$f(sr^{-1}) = f(s)f(r)^{-1} = (12)(132) = (13)$$

which is indeed true. Since the order of $f(r)$ is $3$, and the order of $f(s)$ is $2$, and $f(r)$ / $f(s)$ cannot be produced using each other, the group $X = \left< f(r), f(s) \mid f(r)^3 = f(s)^2 = e, f(s)f(r)^{-1} \right>$ is isomorphic to $D_6$. But $X$ contains $6$ elements and so does $S_3$, so by closure in $S_3$, we must have $X = S_3$.

Are there any general tips for proving isomorphism? Is it a good idea to look at generators and relations between generators, and try to find a correspondence?

Thank you!

## marked as duplicate by Dietrich Burde, Henrik, Pedro Tamaroff♦May 22 '16 at 0:54

• @DietrichBurde While the topics are the same, the questions are very different. I believe the OP here is really hoping to have their proof critiqued (and I have hopefully made this easier by using Mathjax). – pjs36 May 21 '16 at 21:02

A standard way to prove that these two sets are isomorphic is to prove that they satisfy the same defining relations. For this particular example, one can show without too much difficulty (i.e. just write out the full multiplication table) show that

any group of order 6 such that there exist elements $a$ and $b$ where: order of $a$ is 3, order of $b$ is 2, and $ba = a^{-1} b$

is isomporphic to $D_6$.

There is a more general result:

any group of order $2n$ such that there exist elements $a$ and $b$ where: order of $a$ is $n$, order of $b$ is 2, and $ba = a^{-1} b$

is isomorphic to $D_{2n}$.

With a group this small you can almost get away with writing down an isomorphism explicitly as a table and verify by inspection that it is a homomorphism, and is injective and surjective. The only trouble is that there is $36$ different products to check for the homomorphism condition $f(ab)=f(a)f(b)$, and that's at the upper limit of where one would like to do things purely by inspection, if something quicker is possible.

However, since you know a generator-relation representation of $D_6$, you can define a homomorphism $D_6\to S_3$ simply by giving the image of each of the generators. Now the only thing you need in order to show this produces a homomorphism is to show that the images in $S_3$ of the generators satisfy all of the relations between the generators themselves.

Now work out the homomorphism as an explicit table, and use the tabular representation to see by inspection that it is surjective. Since both groups have the same finite order, "surjective" implies "injective", so the homomorphism is an isomorphism!

• Thank you, so is my proof correct? – Ecasx May 21 '16 at 20:26

$D_3$ can be embedded in $S_3$ by sending $r$ to $(123)$ and $s$ to $(23)$ and then extend homomorphically. You can see this intuitively by noticing that any element of $D_3$ is completely determined by its action on the three points of the triangle, and every element of $D_3$ then induces a bijection from this set of points to itself, label the points $1$, $2$ and $3$ and you are there. So $D_3$ is isomorphic to a subgroup of $S_3$. Bur $D_3$ and $S_3$ both have $6$ elements so the subgroup it is isomorphic to must be the whole of $S_3$.

There are four things you need when showing an isomorphism:

$\bullet$ Define a function $\phi: A\rightarrow B$.

$\bullet$ Prove that $\phi$ is injective.

$\bullet$ Prove that $\phi$ is surjective.

$\bullet$ Show that the homomorphism property $\phi(xy)=\phi(x)\phi(y)$ holds.

Generators do map to generators through isomorphism, but that is result of the isomorphism and not a proof. It can help in finding a function $\phi$, however.

Here is another proof, that actually proves a bit more:

Let $G,H$ be two non-abelian groups of order $6$. Then $G \cong H$.

First off, we seek to show each group has an element of order $3$, and an element of order $2$. Since both are non-abelian, we don't have any elements in either of order $6$, for such an element would then generate the entire group, which would then be cyclic, and thus abelian.

So we only have non-identity elements of orders $2$ and/or $3$. Could all the non-identity elements be order $2$? No, because if $x$ and $y$ were two distinct such elements (in either group), then $xy$ would be a third element, by supposition also of order $2$ (Note $xy \neq e$ since $x \neq y$, and $x,y$ are of order $2$). Since:

$e = (xy)^2 = x^2y^2$, we see $x$ and $y$ commute, and thus $\{e,x,y,xy\}$ is a subgroup of $G$ or $H$ of order $4$. But $4\not\mid 6$, so this is impossible.

On the other hand, it is clear that non-identity elements of order $3$ occur in pairs, so it is impossible to have $5$ such. So $G$ (or $H$) has at least one element (and thus at least $2$) of order $3$, and at least one element of order $2$. Let us call these elements $a,b$ (for $G$, and say $a',b'$ for $H$-in what follows, in your mind "put the primes on" to make corresponding statements for $H$)).

Straight off the bat we know $4$ distinct elements: $e,a,a^2,b$. By closure we see that $ab$ is in the group, and cannot equal any of the four prior elements:

$ab = e \implies b = a^{-1} = a^2\\ ab = a \implies b = e\\ ab = a^2 \implies b = a\\ ab = b \implies a = e$

Similarly, we know that $a^2b$ is also in the group, and a similar process of elimination shows it is distinct from the $5$ given so far.

Now $ba$ must also be in the group, and we have just two possibilities: $ba = ab$, or $ba = a^2b$.

If $ba = ab$, then $(ab)^n = a^nb^n$, and thus $ab$ has order $6$. Since $G$ (and $H$) are non-abelian, this cannot be the case.

Thus $G = \{e,a,a^2,b,ab,a^2b\} = \langle a,b: a^3 = b^2 = e, ba = a^2b\rangle$ and similarly:

$H = \langle a',b': a'^3 = b'^2 = e', b'a' = a'^2b'\rangle$ and $\phi:G \to H$ given by:

$e \mapsto e'\\ a \mapsto a'\\ a^2 \mapsto a'^2\\ b \mapsto b'\\ ab \mapsto a'b'\\ a^2b \mapsto a'^2b'$

is the desired isomorphism (it's clearly bijective, and a homomorphism).

So your particular problem comes down to identifying one element of order $3$ in $D_3$ and one element of order $3$ in $S_3$, and similarly one element of order $2$ in each group, and showing they (the element of order $3$, and the element of order $2$) do not commute. The isomorphism you exhibit is (essentially) the same one as in my post above (and if you look closely enough, actually shows we have thereby $4$ possible automorphisms of any non-abelian group of order $6$ (because we might have $G = H$)).