Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to show that $\langle x,y \mid x^3=y^2=(xy)^3=e\rangle$ is isomorphic to $\mathfrak{A}_4$. I've already done most of the proof. The remaining and probably hardest part is to show that $\langle x,y \mid x^3=y^2=(xy)^3=e\rangle$ has at most $12$ elements.

For this I have a hint: Show that the conjugates of $y$, together with the identity element form a normal subgroup of order $\leq 4$ and index $\leq 3$.

I suspect that the conjugates of $y$ are just $y$, $xyx^{-1}$ and $x^2yx^{-2}$, but I really don't know how to prove this. Can somebody help me?

share|cite|improve this question
Hint: Write $y^x$ for $xyx^{-1}$, etc. Then $H = \langle y,y^x \rangle$ is a dihedral group, since it is generated by a pair of elements of order $2.$ Then we have $y.y^{x}.y^{x^{2}} = (yx)^3 =1$ since $yx$ is conjugate to $xy.$ Thus $H = \langle y,y^x,y^{x^{2}} \rangle.$ Then $H$ is normalized both by $y$ and by $x$ – Geoff Robinson Feb 24 '12 at 8:43
@Geoff: Thanks for the hint. I think I can do the proof from here. I'll write an answer then. – Stefan Walter Feb 25 '12 at 11:33
You are welcome! – Geoff Robinson Feb 25 '12 at 11:38
up vote 2 down vote accepted

It is easy to show that $(123)$ and $(12)(34)$ generate $\mathfrak{A}_4$ and satisfy the given relations with $(123)$ in place of $x$ and $(12)(34)$ in place of $y$. So there exists a group homomorphism of $G:=\langle x,y|x^3=y^2=(xy)^3=e\rangle$ onto $\mathfrak{A}_4$. If we can show that $G$ cannot have more than $12$ elements, the proof is finished.

We have $$y(xyx^{-1})(x^2yx^{-2})=(yx)^3x^{-3}=(yx)^3=(y(xy)y^{-1})^3=(xy)^3=e.$$ Since $y$ is equal to its own inverse, the same is true for its conjugates. This information allows us to play around with the equation above to conclude that $H:=\{e,y,xyx^{-1},x^2yx^{-2}\}$ is indeed a subgroup.

There exists a homomorphism $f$ of $G$ onto $\mathbf{Z}/3\mathbf{Z}$ mapping $x$ to $1$ and $y$ to $0$. We show that $\text{Ker}(f)=H$ and thereby $[G:H]=3$, which implies $|G|\leq 12$.

Let $\phi$ be the homomorphism of the free monoid $\text{Mo}(\xi,\eta)$ onto $G$ mapping $\xi$ to $x$ and $\eta$ to $y$. Then $\phi(w)\in\text{Ker}(f)$ if and only if the number of occurences of $\xi$ in $w$ is a multiple of $3$. For $i\in\mathbf{N}$, let $M_i$ be the set of elements of $\text{Mo}(\xi,\eta)$ with $i$ occurences of $\xi$. We show by induction on $n$ that $\phi(M_{3n})\subset H$ for all $n\in\mathbf{N}$.

Obviously, $\phi(M_0)=\{e,y\}\subset H$. If $n\geq 1$ and $w\in M_{3n}$, then there exists $w'\in M_3$ and $w''\in M_{3(n-1)}$ such that $w=w'w''$. Since $\phi(w'')\in H$ by induction, it is left to show that $\phi(M_3)\subset H$. Let $w\in M_3$. Then $w=\eta^i\xi\eta^j\xi\eta^k\xi\eta^l$ with $i,j,k,l\in\mathbf{N}$. So $\phi(w)=y^ixy^jxy^kxy^l$. It remains to show that $xy^jxy^kx\in H$ for all $j,k\in\mathbf{N}$.

There are four cases:

$j$ and $k$ are even $\Rightarrow$ $xy^jxy^kx=x^3=e\in H$.

$j$ is even and $k$ is odd $\Rightarrow$ $xy^jxy^kx=x^2yx=x^2yx^{-2}\in H$

$j$ is odd and $k$ is even $\Rightarrow$ $xy^jxy^kx=xyx^2=xyx^{-1}\in H$

$j$ and $k$ are odd $\Rightarrow$ $xy^jxy^kx=xyxyx=(xy)^3y=y\in H$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.