I do not get prime-decomposition! Please help! So this is a group theory question and I am particularly interested in part b. I  don't see how they are equivalent. I am told that is this is because $C_9 \ncong C_3 \times C_3$, but I was wondering why. Is it that $C_4 \ncong C_2 \times C_2$? If somebody could explain exactly how to break these things down so we can see what is isomorphic to what I would be EXTREMELY grateful! much love. xx

 A: Let's look at $C_2\times C_2$ and $C_4$. 
The four elements of $C_2\times C_2$ are $(0, 0), (1, 0), (0, 1), (1, 1)$. Now let's look at their doubles:


*

*$(0, 0)+(0, 0)=(0+0, 0+0)=(0, 0)$.

*$(1, 0)+(1, 0)=(1+1, 0+0)=(0, 0)$.

*$(0, 1)+(0, 1)=(0+0, 1+1)=(0, 0)$.

*$(1, 1)+(1, 1)=(1+1, 1+1)=(0, 0)$.
That is, every element of $C_2\times C_2$ has order $2$: double it, and you get the identity.
By contrast, let's look at $C_4$, which has four elements: $0, 1, 2, 3$. If we double (say) $1$, we get $2$; this is not the identity! So $C_4$ has elements which are not of order $4$.
This is enough to show that $C_2\times C_2\not\cong C_4$; proving this is a good exercise. HINT: show that if $g\in G$ and $h\in H$ have different orders, then no map sending $g$ to $h$ can be an isomorphism.
Similarly, $C_3\times C_3\not=C_9$. More generally, if $a$ and $b$ are not coprime (= have factors in common; e.g., $4$ and $6$ are not coprime), then $C_a\times C_b\not=C_{a\times b}$. By contrast, if $a$ and $b$ are coprime (= no factors in common; e.g., $15$ and $8$ are coprime), then $C_a\times C_b=C_{a\times b}$. This is another good exercise, and - while significantly more difficult - will reveal a lot about how products of groups behave.
