= Solution
Work with finite <groups>, as required for these <Sylow subgroup> counts. Let $P$ act by <conjugation> on the set of Sylow $p$-subgroups. The <stabilizer> of $Q$ is $P\cap N_G(Q)$. Since $Q$ is the normal <Sylow subgroup> of its own <normalizer>, every $p$-subgroup of that <normalizer> lies in $Q$. Consequently the <stabilizer> is exactly $P\cap Q$, and the <orbit-stabilizer theorem> gives orbit size $[P:P\cap Q]$. The orbit of $P$ has size one. Every other orbit size is a power of $p$ at least $p^a$, hence is divisible by $p^a$. Adding orbit sizes proves the <strengthened Sylow congruence from intersections>:
$$
\boxed{n_p\equiv1\pmod{p^a}.}
$$
Now suppose $G$ is simple of order $2^e\cdot15$. If $e=0$, its Sylow $5$-subgroup is normal because its number divides three and is one modulo five. If $e=1$, an involution in the regular <permutation action> of $G$ swaps fifteen pairs, so its sign is negative. The <sign of a permutation> would give a nontrivial <group homomorphism> $G\to C_2$ with a proper nontrivial normal kernel. Both cases contradict simplicity. Thus $e\ge2$.
The <Sylow theorems> make $n_2$ a divisor of fifteen. It is not one, since a nontrivial <Sylow subgroup> would then be normal. It is not three, since the <conjugation action> on three <Sylow subgroups> would give a faithful <group homomorphism> $G\hookrightarrow S_3$, impossible by order. If $n_2=5$, take a <Sylow subgroup> $P$ itself: its <normalizer> has <subgroup index> five. If $n_2=15$, not all distinct <Sylow subgroups> can intersect $P$ in <subgroup index> at least four, since that would give $15\equiv1\pmod4$. Choose $Q\ne P$ with $[P:P\cap Q]=2$. Put $R=P\cap Q$, of order $2^{e-1}$. It has <subgroup index> two in both $P,Q$, so is normal in each. Hence $N_G(R)$ contains both distinct <Sylow subgroups> and has order greater than $2^e$. It is proper: otherwise $R$ would be a nontrivial proper <normal subgroup> of $G$. Its order is therefore $2^e\cdot3$ or $2^e\cdot5$. The latter would give an index-three subgroup and again an impossible faithful action on three <cosets>. Thus $[G:N_G(R)]=5$.
In either case we have the requested subgroup of order $2^e$ or $2^{e-1}$ with index-five <normalizer>. The action on the five <cosets> of that <normalizer> is nontrivial and faithful by simplicity. Its image lies in $A_5$, since composing with sign cannot give a nontrivial map from this <simple group> of order greater than two to $C_2$. Thus $2^e\cdot15$ divides $60$, forcing $e=2$ and equality of orders. This proves the <simple groups with order a power of two times fifteen> conclusion:
$$
\boxed{G\cong A_5.}
$$
For $SL_2(5)$, choose the first column: there are $5^2-1=24$ nonzero choices. For each, determinant one is a nonzero linear equation in the second column and has five solutions. Therefore $|SL_2(5)|=120$. A central <matrix> commuting with $\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ has zero lower-left entry and equal diagonal entries; commuting also with $\left(\begin{smallmatrix}1&0\\1&1\end{smallmatrix}\right)$ makes the upper-right entry zero. It is scalar. Determinant one then gives
$$
\boxed{|SL_2(5)|=120,\qquad Z=\{I,-I\},\quad |Z|=2.}
$$
To establish the <projective special linear group over the field with five elements> isomorphism without assuming its simplicity, compute its classes. For determinant-one matrices, the <characteristic polynomial> is $X^2-tX+1$ with trace $t$. The possibilities over $\mathbb F_5$ exhaust the group:
$$
\begin{array}{c|c|c}
\text{type in }SL_2(5)&\text{centralizer order}&\text{class sizes in }SL_2(5)\\\hline
\pm I&120&1,1\\
t=0&4&30\\
t=1,-1&6&20,20\\
t=2,-2,\ \text{noncentral}&10&12,12,12,12
\end{array}
$$
For trace zero the <eigenvalues> are $2,3$, so the determinant-one <centralizer> is the split diagonal torus of order four. For traces $\pm1$, the discriminant is the nonsquare $2$: the <centralizer> is the norm-one subgroup of $\mathbb F_{25}^{\times}$, of order six. In both semisimple cases the determinant map on the full linear <centralizer> is surjective onto $\mathbb F_5^{\times}$; consequently each trace type is a single special-linear class. For the noncentral traces $\pm2$, a <Jordan block> has full linear <centralizer> $\left\{\left(\begin{smallmatrix}a&b\\0&a\end{smallmatrix}\right):a\ne0\right\}$ with determinant $a^2$. The <determinant> image consists of the two squares, so each general-linear class splits into two special-linear classes; the determinant-one <centralizer> has order $2\cdot5=10$. These sizes sum to $120$.
Quotienting by $Z$ identifies a matrix with its negative. The trace-zero class gives one projective class of size fifteen; the two traces $\pm1$ give one class of size twenty; the four noncentral unipotent classes give two classes of size twelve. Their respective projective orders are $2,3,5,5$, by the <characteristic polynomial>s. A <normal subgroup> is a union of classes containing the identity. The nontrivial proper possible union sizes are $13,16,21,25,28,33,36,40,45,48$, none of which divides $60$. Thus the quotient is simple. Applying the order-$60$ result above gives
$$
\boxed{PSL_2(5)\cong A_5.}
$$
Finally, if $A^2=I$ over an odd-characteristic field, its <minimal polynomial> divides the square-free polynomial $(X-1)(X+1)$, so it is diagonalizable with <eigenvalues> $\pm1$. In dimension two, determinant one requires both eigenvalues equal. Hence $A=I$ or $A=-I$. This proves the <unique involution in SL2 over an odd field> assertion: \b[the only element of order two in $SL_2(5)$ is $-I$]. If an index-two subgroup $H$ existed, it would be normal and have order sixty. By <Cauchy theorem> it would contain an involution, hence contain $Z$. Then $H/Z$ would be a <normal subgroup> of order thirty in the simple quotient $PSL_2(5)$, a contradiction. Thus \b[$SL_2(5)$ has no index-two subgroup].
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