I'm studying the symmetric group $S_n$ of permutations from $\{1,2,\ldots,n\}$ to itself. I'm doing a problem where I have to show that $$S_4\cong \langle a,b~|~a^4=b^2=(ba)^3=1\rangle.$$ I thought the following: in general, a $k$-cycle has order $k$ in $S_n$ for $k\leq n$ (we can think about $\pmod n$ but I want to simplify it).
On the other hand, I've alredy proved that $S_n=\langle (1~2),(1~2\cdots~n)\rangle$, so $\langle(1~2)(1~2~3~4)\rangle S_4$.
Let $a:=(1~2~3~4),~b=(1~2)$.
Indeed, $a^4=(1~2~3~4)^4=1$, $b^2=(1~2)^2=1$, and I was checking that third condition, $(ba)^3=1$.
Doing the product of cycles, I got that $(1~2)(1~2~3~4)=(2~3~4)$.
However, I'm surely making a big mistake when checking with an example. Consider the set $(1,2,3,4)$ (which is a $4$-tuple). Then, applying $a$ to that set it becomes $\color{red}{\text{NOT}}$ $\color{red}{(4,1,2,3)}$, $(2,3,4,1)$ and then $(1,2)$, it will be $(1,3,4,2)$.
But when I apply the cycle $(2~3~4)$ to the $4$-tuple $(1,2,3,4)$, $1$ is fixed and they're alredy different. In fact I get $(1,3,4,2)$ which is different to $(4,2,1,3)$.
What am I doing wrong here? I want the product $ba$ so then I compute $(ba)^3$ and check that it is indeed $1$ (and in this case, $(2~3~4)^3=1$).
Edit: red text is what I was doing wrong and now changing it with help of the comments.
Edit 2: Solved!!