= Solution
A <simple group> has no <normal subgroup> except the identity <subgroup> and the whole group. To prove the <Simplicity of the alternating group A5>, we compute its <conjugacy classes>. Since the <sign homomorphism> is onto, $|A_5|=5!/2=60$. Its possible nonidentity <cycle types> are a three-cycle, a product of two disjoint transpositions, and a five-cycle.
There are $\binom53\cdot2=20$ three-cycles. The <centralizer> in $S_5$ of $(1\ 2\ 3)$ consists of its three powers and an independent interchange of the two unused letters. Only the three powers are even. Its <centralizer> in $A_5$ therefore has order three, giving one <conjugacy class> of size $60/3=20$ by the <orbit-stabilizer theorem>.
There are $5\cdot3=15$ products of two disjoint transpositions: choose the fixed letter and then a pairing of the other four letters. For $(1\ 2)(3\ 4)$, a commuting <permutation> fixes $5$ and may interchange the two pairs and swap within them, giving a <centralizer> of order eight in $S_5$. It contains an odd element, $(1\ 2)$, so exactly half its elements are even. The <centralizer> in $A_5$ has order four, and these fifteen elements form one <conjugacy class> of size $60/4=15$.
Finally, there are $4!=24$ five-cycles. A <permutation> commuting with a five-cycle is determined by the image of one letter and must be a power of that cycle. All five powers are even, so the <centralizer> in $A_5$ has order five. Each such <conjugacy class> has size $60/5=12$, giving two classes of size twelve.
A <normal subgroup> must be a union of <conjugacy classes> and must include the identity. Its order therefore has the form
$$
1+20a+15b+12c,\qquad a,b\in\{0,1\},\quad c\in\{0,1,2\}.
$$
The resulting distinct possibilities are
$$
1,13,16,21,25,28,33,36,40,45,48,60.
$$
Of these, only $1$ and $60$ divide $60$. <Lagrange's theorem> rules out every other possible order. Hence
$$
\boxed{A_5\text{ is simple}.}
$$
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