I'm currently on the chapter of internal direct product in group theory, and became curious about the following questions:

  1. One definition of an internal direct product in my book is that for a group $G$ with $S_i$ being a subgroup of $G$ and $G=S_1\cdots S_n$, $\prod_{i=1}^nS_i \cong G$ canonically by $h(s_1,\cdots,s_n)=s_1\cdots s_n$. My question is, is it necessary to restrict to the canonical isomorphism? I tried to find an example where $\prod_{i=1}^nS_i \cong G$ only non-canonically, but I couldn't.

  2. I've learned that a finitely generated Abelian group can be decomposed into cyclic groups $G=C_1\oplus \cdots \oplus C_n$ such that $C_{i\leq k}$ is of order $m_i$ with $m_1|m_2|\cdots|m_k$ and $C_{i>k}$ is infinite. My question is, is this decomposition unique in any sense? In particular,

(1) For a finite Abelian group, each $m_i$ is uniquely determined. But when the group is not cyclic, the converse of Lagrange's theorem doesn't hold in general, so I guess $G=D_1\oplus \cdots\oplus D_n$ is possible where $D_i\neq C_i$ for some $i$ (even though they are isomorphic since they are cyclic group of same order).

(2) When $G$ is infinite, since my book proved that $m_i$s are uniquely determined only for a finite group, I think the same assertion wouldn't necessarily hold in $G$. The problem is, I'm just a beginner and haven't met with infinite group other than cyclic group so I can't come up with suitable example of such a group, i.e. $G=\bigoplus_{i=1}^l D_i$ with possibly $l\neq n$ and the finite part of some $D_i$ has different order than $m_i$.

Sorry if this question was basic, but I was so curious about this. It would be grateful if there is an example I can understand; I started learning group theory this semester.

  • $\begingroup$ That first definition makes no sense. You can just take any finite number of subgroups, take the product and have an isomorphism...no. Shouldn't they be normal? Intersection trivial? Product actually G? That it the right map but you need assumptions other that what you have. Also, for your second question look at invariant factor decomposition (which you described), this will tell you that this is unique. $\endgroup$ – Mr.Fry Jun 4 '15 at 0:08
  • $\begingroup$ Sorry, I changed the definition correctly. $\endgroup$ – Taxxi Jun 4 '15 at 0:29
  • $\begingroup$ Come to think of it, isn't isomorphism required to be bijective? I think $G=S_1\cdots S_n$ is not needed. $\endgroup$ – Taxxi Jun 4 '15 at 0:48
  • $\begingroup$ Yes. Iso = hom + bijection. The product is needed, look at the map. You say $g = h_1 \cdots h_n$ and define the map $h_1 \cdots h_n \mapsto (h_1,...,h_n)$. $\endgroup$ – Mr.Fry Jun 4 '15 at 0:50
  1. http://groupprops.subwiki.org/wiki/Internal_direct_product

    Example: Show $\mathbb{Z}_6 \cong \mathbb{Z}_3 \oplus \mathbb{Z}_2$

  2. http://www.millersville.edu/~bikenaga/abstract-algebra-1/fg-abelian-groups/fg-abelian-groups.html

    Example: Let $|G| = 2^3 \cdot 3^2 \cdot 5$. Write down all nonisomorphic abelian groups

  • $\begingroup$ Thanks for the first simple example. But for the second case, I do know how to decompose it 'up to isomorphism', but my question (2-1) was about components of a direct product decomposition (as a subgroup). And for (2-2), invariant factor decomposition will indeed answer my question, but my book didn't define 'invariants' for an infinite yet finitely generated Abelian group (and the proof for invariant is restricted for finite group only). I think a general proof requires the concept of ring or several others as well unfortunately... Thanks for it anyway; have to study more. $\endgroup$ – Taxxi Jun 4 '15 at 0:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.