We know that if $X_1 \sim \hbox{Normal}(\mu_1, \sigma^2_1)$ and $X_2 \sim \hbox{Normal}(\mu_2, \sigma^2_2)$, then: $$X = X_1+X_2 \sim \hbox{Normal}\left(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2\right)$$
But how about a random sum? That is, consider $X_j \overset{iid}{\sim} \hbox{Normal}(0,\sigma^2)$ and $N \sim \hbox{Poisson} (\lambda)$ independent of $(X_j)_{j=1}^{\infty}$. So, define $$X = X_1 + X_2 + \ldots + X_N$$ What is the distribution of $X$?
It is straightforward to show that $E[X]=0$ and $\mathbf{var}(X)= \lambda \mathbf{var}(X_j)=\lambda \sigma^2$. So I think that $X \sim \hbox{Normal}(0,\lambda \sigma^2)$. But I can't show this.
The claim is false. $X$ has a compound Poisson distribution with characteristic function $$E[e^{i\xi(X_1+...+X_N)}]=e^{\lambda(e^{-\xi^2\sigma^2/2}-1)}$$ which is not a normal characteristic function.