# A problem in decomposing a p group into direct sum of nontrivial subgroups

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 $b=n-a<n$.

• wow what a fast and helpful response much obliged – Don John Prep Dec 30 '15 at 22:07
• Wow, I only found this question like 5 minutes after it was posted, I didn't think my answer was so fast. If you can mark my answer as correct (the green checkmark), I'd be grateful. – Alex Dec 30 '15 at 22:09

With $\mathbb{Z}/(p^n)$ is meant the quotient group. More precisely it is the set $\{k+(p^n): k\in \mathbb{Z}\}$ endowed with the group structure given by $(k+(p^n))+(k'+(p^n))=(k+k')+(p^n)$.

Showing that a group is indecomposable is the same as showing that the any idempotent of $End(G)$ is trivial or the identity. In this vein let $f$ be any idempotent and let $f(1)=k$. Then $k=f(1)=f^2(1)=f(f(1))=f(k)=kf(1)=k^2$. Thus $k(k-1)=0$. So $p^n$ divides $k(k-1)$ in $\mathbb{Z}$. But $k$ and $k-1$ are coprime, so $p^n$ divides $k$ or $k-1$, i.e. $k=0$ or $k=1$ in $\mathbb{Z}/(p^n)$.

Edit: $f\in End(G)$ is an idempotent if $f^2=f$. Any idempotent $f$ gives a decomposition $G=Im(G) \oplus Ker(g)$ and a decomposition $G=H \oplus I$ gives an idempotent by projecting onto the first coordinate. These two constructions are mutually invers. So we get a bijection {direct decompositions of G} $\longleftrightarrow$ {idempotents of $G$}

• Thank you for this great answer – Don John Prep Dec 30 '15 at 22:10
• I do not understand what idempotent means – Don John Prep Dec 30 '15 at 22:14
• i edited the answer. – math635 Dec 30 '15 at 22:21