= Solution
{c}
First justify that the <moments> being used exist. For every $r>0$,
$$
\mathbb E e^{r|Y|}\leq\mathbb E e^{rY}+\mathbb E e^{-rY}=M_Y(r)+M_Y(-r)<\infty.
$$
The bound $|y|^k\leq k!r^{-k}e^{r|y|}$ gives $\mathbb E|Y|^k<\infty$ for all $k$.
If $Y$ has a <symmetric distribution>, then $Y$ and $-Y$ have the same <probability distribution>, so for odd $k$,
$$
\mathbb EY^k=\mathbb E(-Y)^k=-\mathbb EY^k,
$$
which forces $\mathbb EY^k=0$.
Conversely, suppose every odd <moment> vanishes. For fixed $t\in\mathbb R$, the absolute values of the partial <Taylor series> for $e^{tY}$ are bounded by $e^{|t||Y|}$, whose <expected value> is finite by the preceding inequality. The <dominated convergence theorem> therefore permits termwise <expectation>:
$$
M_Y(t)=\sum_{k=0}^\infty\frac{t^k\mathbb EY^k}{k!}
=\sum_{j=0}^\infty\frac{t^{2j}\mathbb EY^{2j}}{(2j)!}=M_Y(-t)=M_{-Y}(t).
$$
The <uniqueness theorem for moment-generating functions> implies equality of the <probability distributions> of $Y$ and $-Y$. Hence
$$
\boxed{Y\text{ has a symmetric distribution}\ \Longleftrightarrow\ \mathbb EY^k=0\text{ for every odd }k.}
$$
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