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 $A_4$.