The definition of a primecomponent of an abelian group $G$ for a prime $p$ is the following: $$ G_p:\{a\in G: \text{ord} (a) = p^{l(a)} \}<G $$

Now the theorem about the decomposition of finite abelian groups in primecomponents says that with $G$ abelian finite group and ord$(G)=p_1^{k_1}\cdot...\cdot p_1^{k_r}$ with pairwise different $p_1,...,p_r$ then $$G=G_{p_1}\times ...\times G_{p_r}$$ is an inner direct product and ord$(G_{p_i})=p_i^{k_i}$ for $i=1,...r$.

I think I just struggle with the notation. All along here I will write $Z(n)=Z_n=\mathbb{Z}/n\mathbb{Z}$. It is clear to me that from the theorem of the prime decomposition we can group together all the factors of the same prime number, so for example if the prime decomposition has that $$ G\cong Z(q_1^{l_1}) \times Z(q_1^{l_2}) \times ... \times Z(q_1^{l_s}) $$ and $p=q_1,...q_n$ and $p\neq q_i$ for $i\in\{n+1,...,s\}$ we could write $$ G_p=Z(p^{l_1}) \times Z(p^{l_2}) \times ... \times Z(p^{l_n}) $$ and I would see how because $G_p$ are subgroups of an finite abelian group there are normal, $G_{p_1}\cdot...\cdot G_{p_r}=G$ and also disjunct, so it is an inner product. But what I cannot understand is how this defined $G_p$ is a subgroup of G. For example $$ G:= Z(2) \times Z(2^2)\times Z(2^4)\times Z(3)\times Z(3^3)\times Z(5) $$ One would assume $G_2=Z(2) \times Z(2^2)\times Z(2^4)$, $G_3=Z(3)\times Z(3^3)$ and $G_5=Z(5)$. I just can't understand how $G_2<G$ is a subgroup of G. For example let $(a_1,a_2,a_3) \in Z(2) \times Z(2^2)\times Z(2^4)$. But how is $(a_1,a_2,a_3)$ also in the much larger product $Z(2) \times Z(2^2)\times Z(2^4)\times Z(3)\times Z(3^3)\times Z(5)$? Are elements of $G$ not of the form $(a_1,a_2,a_3,a_4,a_5,a_6)$?


But $G_1\times e\le G_1\times G_2$ quite clearly, in general.

  • $\begingroup$ Yes, that's quite clear of course. But then one would get into trouble with the $G_i$ being disjoint for the conclusion of the inner direct product, they have to be normal, their product has to be $G$ and they need to be disjoint right? $\endgroup$ – KingDingeling Jan 20 '20 at 11:22
  • $\begingroup$ Disjoint? Well, $G\times e\cap e\times G=(e,e)=e_{G\times G}$. Consider the Klein four group. You can cross a group with itself. $\endgroup$ – Chris Custer Jan 20 '20 at 11:29
  • $\begingroup$ The factors of the product automatically have trivial intersection. $\endgroup$ – Chris Custer Jan 20 '20 at 11:36
  • 1
    $\begingroup$ Yes well... this explains it. Thank you very much. So I have to understand $G_2=Z(2) \times Z(2^2) \times Z(2^4) \times \{e\} \times \{e\} \times \{e\}$ right? $\endgroup$ – KingDingeling Jan 20 '20 at 11:42
  • 1
    $\begingroup$ Bingo. You are correct. $\endgroup$ – Chris Custer Jan 20 '20 at 11:47

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.