Question is to prove that :

Subgroup of $A_4$ generated by an element of order $2$ and an element of order $3$ is all of $A_4$.

what i have done so far is.

Taking $H=\big< a,b\big>\leq A_4$ with $a^2=b^3=1$, I see that $H$ has atleast $6$ elements $1,a,b,b^2,ab,ba$

By lagrange theorem there can be no proper subgroup in between $H$ and $A_4$ and we know that $A_4$ has no subgroup of order $6$.

Thus $H$ must be equal to whole group $A_4$ i.e., $A_4$ is generated by any element of order $2$ and an element of order $3$.

I remember the result that $A_4$ has no subgroup of order $6$ But i do not remember the proof and i am unable to reproduce immediately.

I would be thankful if some one can verify the way i am done and give a hint for my assumption.

P.S : I have tried something on the way to conclude that "$A_4$ has no subgroup of order $6$ "

Suppose $H\leq A_4$ with order 6, then $H\unlhd A_4$. As $G/H=2$ for any element $g\in A_4$, we have $g^2\in H$ and then i see that i have 9 elements in $A_4$ which are squares.But, H has only 6 elements. So, i see that $H$ is not a subgroup of $A_4$.

  • $\begingroup$ that was neglected for simple reason.suppose $ab=ba$ then, $ab$ has order $6$ and then in H, i will have one element (from $<a>$) two elements from $<b>$ and 5 elements from $<ab>$ adding up to $8$ which would trivially imply it should be the whole group by lagrange theorem. $\endgroup$ – user87543 Sep 2 '13 at 14:02

You need two facts to prove that $A_4$ does not have a subgroup of order 6 :

(a) If $H<G$ such that $[G:H] = 2$, then $H$ is normal in $G$.

Proof : Just show that, for any $a \notin H$, $aH = G\setminus H = Ha$.

(b) If $H\vartriangleleft G$, and $g\in G$ is such that $o(g)$ is relatively prime to $[G:H]$, then $g \in H$.

Proof : Just look at the order of $\overline{g}$ in $G/H$. It must divide both $o(g)$ and $[G:H]$.

Now, if $A_4$ had a subgroup $H$ of order 6, then by (a), it would be normal in $A_4$. By (b), $H$ must contain all elements of odd order. There are 8 elements in $A_4$ of order 3 (the 3-cycles), and so $|H| > 8$, which is a contradiction.

  • $\begingroup$ I have edited my question, please have a look at that. $\endgroup$ – user87543 Sep 2 '13 at 14:00

I think if we focus on the order of $ab$, some valuable things will happen. Clearly $$|ab|=2,3,12$$ It can't be that $|ab|=12$ and if $|ab|=2$ then the following subgroup would exist: $$\langle a,b\mid a^2=b^3=(ab)^2=1\rangle\cong S_3$$ But as you may consult many text like here; $A_4$ has no any subgroup of order $6$. So $|ab|=3$. This means that $$a^2=b^3=(ab)^3=1$$ Indeed:

$$\langle a,b\mid a^2=b^3=(ab)^3=1\rangle$$ is a nice presentation of $A_4$.

  • $\begingroup$ Nicely done. +1 $\endgroup$ – Namaste Sep 3 '13 at 11:52

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