= Solution
{c}
Label the vertices of a <regular tetrahedron> $1,2,3,4$, and include all its <Euclidean isometries>, both <rotations> and <reflections>. The following list describes every <subgroup> of <cardinality> at most six, allowing arbitrary relabelling of the vertices. The numbers count actual <subgroups>, rather than just their <group isomorphism> types.
* \b[One trivial <subgroup>] contains only the <identity element>.
* \b[Six reflection <subgroups> $C_2$] are generated by a <transposition>, for example $(12)$. Its mirror plane contains vertices $3,4$ and the midpoint of the opposite edge $12$.
* \b[Three half-turn <subgroups> $C_2$] are generated by a product of two disjoint <transpositions>, for example $(12)(34)$. The <rotation> axis joins the midpoints of the opposite edges $12$ and $34$.
* \b[Four <cyclic subgroups> $C_3$] consist of the three <rotations> about an axis joining a vertex to the centre of its opposite face. For example, $\langle(123)\rangle$ fixes vertex $4$.
* \b[Three <cyclic subgroups> $C_4$] are generated by an <improper orthogonal transformation>: a quarter-turn about an axis through opposite edge midpoints, composed with a <reflection> in the perpendicular plane through the centre. For $\langle(1234)\rangle$, the axis joins the midpoints of edges $13$ and $24$. Its square is the half-turn $(13)(24)$.
* \b[One rotational <Klein four-group>] consists of the <identity element> and the three half-turns about opposite-edge midpoint axes:
$$
\{e,(12)(34),(13)(24),(14)(23)\}.
$$
* \b[Three further <Klein four-groups>] arise from a pair of opposite edges. For edges $12,34$, the <subgroup> is
$$
\{e,(12),(34),(12)(34)\}.
$$
It contains two <reflections> in perpendicular planes and their product, the half-turn about the line where those planes meet. It preserves each of the two opposite edges as a <set>.
* \b[Four <subgroups> $S_3$] are the <stabilizer subgroups> of individual vertices. Each contains three <rotations> about the vertex-to-opposite-face axis and three <reflections> in planes containing that axis. For the fixed vertex $4$, this is every <permutation> of $1,2,3$.
Here is why this is exhaustive. Part <6/b> leaves only the <trivial group>, $C_2,C_3,C_4,C_2\times C_2,S_3$. The <disjoint cycle decomposition> gives exactly the two listed kinds of <involutions>, the three-cycles, and the four-cycles, accounting for all the <cyclic subgroups>. In a <Klein four-group>, the nonidentity elements commute. If it contains $(12)$, any other nonidentity element must be a commuting <involution>, hence $(34)$ or $(12)(34)$; either choice yields the displayed opposite-edge <subgroup>. If it has no <transposition>, its three nonidentity elements must be all three double <transpositions>, yielding the rotational <subgroup>.
Finally, an $S_3$ <subgroup> has a unique <cyclic subgroup> of <cardinality> three. Its two three-cycles fix the same unique vertex. Every element of the $S_3$ <subgroup> preserves this vertex, since <conjugation> preserves that unique <cyclic subgroup>. Thus it is the full <stabilizer subgroup> of that vertex. The counts in the list are consequently exhaustive, giving
$$
\boxed{1+6+3+4+3+1+3+4=25\text{ subgroups of order at most }6.}
$$
The two kinds of $C_2$ and the two kinds of <Klein four-groups> have different geometric behavior despite having the same abstract <group isomorphism> type within each pair. This is the distinction recorded by the <small subgroups of tetrahedral symmetry>.
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