= Solution
In additive notation every element $(a,b)\in\mathbb Z_2\times\mathbb Z_4$ obeys $4(a,b)=(0,0)$. Its <order of a group element> therefore divides four, whereas $1\in\mathbb Z_8$ has order eight. A <group isomorphism> preserves element orders, so
$$
\boxed{\mathbb Z_2\times\mathbb Z_4\not\cong\mathbb Z_8.}
$$
For the second product, define the <group homomorphism>
$$
\Psi:\mathbb Z_6\longrightarrow\mathbb Z_2\times\mathbb Z_3,\qquad
[t]_6\longmapsto([t]_2,[t]_3).
$$
It is well-defined because multiples of six are divisible by both two and three. Its <kernel> contains only zero: simultaneous divisibility by two and three implies divisibility by six. Thus $\Psi$ is injective, and equal finite <group orders> make it surjective. Explicitly the six images are
$$
(0,0),(1,1),(0,2),(1,0),(0,1),(1,2).
$$
Therefore
$$
\boxed{\mathbb Z_2\times\mathbb Z_3\cong\mathbb Z_6.}
$$
For general positive integers $n,m$, the <order of a direct-product element> is the <least common multiple> of its component orders. Hence every element order divides $\operatorname{lcm}(n,m)=nm/\gcd(n,m)$, while $(1,1)$ attains this order. It generates the whole product precisely when $n,m$ are <coprime>:
$$
\boxed{\mathbb Z_n\times\mathbb Z_m\cong\mathbb Z_{nm}\quad\Longleftrightarrow\quad\gcd(n,m)=1.}
$$
This is the <cyclicity of a product of two finite cyclic groups>, with the explicit isomorphism furnished by the <Chinese remainder theorem>.
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