Hello all I have taken a group theory course where we are now covering p groups and we I have met the following exercise:
Let $ G = Z/(p^n) $ is a(n Abelian) group of order $ p^n $ for a prime $ p $ and a natural $ n $.
We are asked to show G cannot be written as a direct sum of two non trivial proper subgroups
Now, first of all what does $ Z/(p^n) $ mean exactly? I have not met this notation previously.
And finally to prove the result, I have found a link on here with a similar problem where the proposed (and accepted) answer was to assume to get contradiction we could do such a thing but then the two non trivial subgroups would intersect nontrivially and the answer stated that for this group this cannot happen which I do not understand how to show. I certainly appreciate all kind helpers willing to assist a novice, thanks
$\Bbb Z/p^n$ is essentially all numbers from $0$ to $p^n-1$, that you can add and multiply; and when you hit a number outside the range of $0$ to $p^n-1$ then you must add/subtract multiples of $p^n$ to put it into that range.
Suppose that it is a product of two non-trivial proper subgroups, say there is an isomorphism:
$\phi: \Bbb Z/p^n \cong \Bbb Z/p^a \oplus \Bbb Z/p^b$, without loss of generality $0<a<b$. By computing the number of elements on both sides, we have $a+b = n$.
Take $\phi(1)$. Then it is an element of order $\leq p^b$, since $p^b * (c, d) = (p^a(p^{b-a}c), p^bd) = (0p^{b-a}c, 0d)=(0, 0)$ for any $(c, d)\in\Bbb Z/p^a \oplus \Bbb Z/p^b$.
This means that $\phi(p^b) = p^b\phi(1)=(0, 0) = \phi(0)$. Since $\phi$ is an isomorphism, then $p^b = 0$, a contradiction since $b=n-a<n$.