Here :
https://groupprops.subwiki.org/wiki/Groups_of_order_16
I noticed the presentation of the $\ 9$-th group with order $16$.
Is this presentation actually correct or a typo ?
I think, it should be $\ a^8\ $ instead of $ a^4\ $. but maybe I miss something.
For the benefit of those who don’t want to go wade through another website, you are talking about the presentation of the generalized quaternion group of order 16, given as $$Q_{16} = \langle a,b\mid a^4=b^2=abab\rangle.\tag{1}$$
There is no error or omission in this description; the only convention at play is that the equalities asserted (plus any that can be deduced from them) are the only ones assumed. There is no implicity “$=e$” when you give this type of presentation.
Note that from $b^2=abab$ we conclude that $b=aba$, so $ab=ba^{-1}$ and $ba=a^{-1}b$. Therefore, $bab^{-1} = a^{-1}$. Therefore, $b^2=b(b^2)b^{-1} = ba^4b^{-1} = (bab^{-1})^4 = a^{-4}$. But we also know $b^2=a^4$, so $a^4=a^{-4}$, hence $a^8=e$. Thus, $a$ has order dividing $8$, which in turn gives that $b^4 = (b^2)^2 = (a^4)^2 = e$, and $b$ has order dividing $4$.
A more common presentation (e.g., from Leedham-Green and McKay’sThe Structure of Groups of Prime Power Order, page 28) is $$Q_{16} = \langle x,y\mid y^8 = 1, x^2=y^{4}, y^x=y^{-1}\rangle.\tag{2}$$ To verify that (1) and (2) give the same presentation, we have already established that $a^8=1$, $(b^{-1})^2=a^{4}$, and $a^{b^{-1}} = bab^{-1}=a^{-1}$, so there is a map from (2) to (1) by sending $y$ to $a$ and $x$ to $b^{-1}$. Conversely, we can define a morphism from (1) to (2) by mapping $a$ to $y$ and $b$ to $x^{-1}$, because $y^4=(x^{-1})^2$ (since $x^4=1$, so $x^2=x^{-2}$). Finally, we need to verify that $y^4=x^2=yx^{-1}yx^{-1}$. From $x^{-1}yx=y^{-1}$ we get $yx=xy^{-1}$, so $yx^{-1} = x^{-1}y^{-1}$. Therefore, $yx^{-1}yx^{-1} = x^{-1}y^{-1}yx^{-1} = x^{-2}=x^2$. Thus, $y$ and $x^{-1}$ satisfy the relations of $a$ and $b$, and we get a morphism from (2) to (1).
The two maps are clearly inverses of each other, as they are inverses on the set of generators, so this shows the two presentations define the same group, namely $Q_{16}$.