Besides to @Betty's points, there is another way for seeing why does this happen. We know that $S_4$ can have the following presentation:
$$S_4=\langle a,b\mid a^2=b^4=(ab)^3=1\rangle$$ Let's satisfy $a=(1,2),~~b=(1,2,3,4)$ in above relations. Indeed $a$ and $b$ can do that, but what will happen if we set $a=(1,3),~~b=(1,2,3,4)$? By this assumption, we see that $(ab)^3=(1,4)(2,3)$ and this happens cause of the points @Betty indicated them in detailed. Now if you are familiar to one of $D_8$'s presentation, then you'll have $$D_8=\langle a,b\mid a^2=b^4=(ab)^3=1\rangle,~~a=(1,3),~~b=(1,2,3,4)$$ instead wich is of order $8$.