Whether there exists or not, a group of $2$ elements $a$ and $b$ with $\lvert b \rvert = 2 , \lvert a\rvert = 15$ and ... Does there exist a group with two elements $a$ and $b$ with $\lvert b\rvert = 2, \lvert a\rvert = 15$  and $bab = a^4$?
 A: Yes. There is such a group. It is a semi-direct product of cyclic groups of respective orders 15 and 2. The reason why this works is that the mapping $x\mapsto x^4$ is an automorphism
of order 2 of the cyclic group generated by $a$. Therefore we have a homomorphism from $C_2$
to $\mathrm{Aut}(C_{15})$, and can build the semi-direct product in the usual way.
If you want a concrete version of such a group, here it is as a subgroup of $S_{15}$. Let $a$ be the 15-cycle (here $A=10,B=11,\ldots$)
$$
a=(123456789ABCDEF),
$$
so $a^4=(159D26AE37BF48C)$.
Then let $b$ be the product of six disjoint 2-cycles
$$
b=(25)(39)(4D)(7A)(8E)(CF).
$$
The usual conjugation trick of permutations then shows that
$$
bab^{-1}=bab=a^4.
$$
The group has 30 elements. They are either of the form $a^i$ or of the form $ba^i$, $0\le i<15$.

To make sure. We don't have too many choices for the integer $m$ in the relation
$bab^{-1}=a^m$. This is because it has to fit together with the fact that $b^2=1$,
so $b^2$ has to commute with $a$.
The following calculation uses this
$$
a=1a1=b^2ab^{-2}=b(bab^{-1})b^{-1}=ba^mb^{-1}=(bab^{-1})^m=(a^m)^m=a^{m^2}.
$$
For this to hold, we must have $m^2\equiv1\pmod{15}$. This leaves are four choices (pairwise non-congruent modulo $15$): 
$m=1$, $m=-1$, $m=4$ and $m=-4$. Of these, the first choice leads to the direct product
$C_{15}\times C_2$, the choice $m=-1$ gives us the dihedral group $D_{15}$ of 30 elements.
The third choice gives the group in question.
A: The group generated by $a, b$ with $a$ of order 15, $b$ of order 2, and $bab=a^4$ is a group of order 30. Any element can be written as a power of $a$ followed by a power of $b$.
The subgroup $H$ generated by $a^3$ is of order 5 and normal: $ba^3b=(bab)^3=(a^4)^3=a^{-3}$.
The subgroup $K$ generated by $a^5$ and $b$ is abelian of order 6: $ba^5b=(a^4)^5=a^5$
Also $G=HK$ but the group is not the direct product. The center is cyclic of order 3 generated by $a^5$.
A: You could also manually give the presentation $G = \langle a, b | a^{15} = b^2 = 1, bab = a^4 \rangle$
We might ask, how big is this group? I claim that it's not so hard to see that this group has at least $30$ elements. Does it have more?
Note that $bab = a^4 \iff a = ba^4 b$ as $b^{-1} = b$. So given any word, we can use this to simplify the word quite a bit. In particular, we need to know how to simplify words that look like $ba^3ba^2ba^5$, some alternating set of $b$ and powers of $a$. But an $a$ will be sandwiched between two $b$ terms in such a word. In this case, I might take the $a^2$ term and write it as $(ba^4b)(ba^4b) = ba^8b$. In general, $a^n = ba^{4n}b$.
Thus $ba^3ba^2ba^5 = b(ba^{12}b)ba^2ba^5 = a^{14}ba^5= a^{14}b(ba^{20}b) = a^{14}a^{20}b = a^4b$, for example.
More directly, we might also want to see that $ba^j = b(ba^{4j}b) = a^{4j}b$, so that we can write any two term word in the form $a^n b$. Combining these ideas together (more formally that I do here), we see that the only words that are left are $\{1, a, a^2 , \ldots a^{14}, b, ab, a^2b, \ldots, a^{14}b\}$, and so this is a group of order $30$. (In fact, I believe this is the same group as in the other answer, but with a much less clever presentation).
