# Finite abelian groups and subgroups.

Let $G$ be a finite abelian group of order $n=p_1^{a_1}\cdot \cdot \cdot p_k^{a_k}$ and $H$ a subgroup of $G$ of order $m=p_1^{b_1}\cdot \cdot \cdot p_k^{b_k}$.

By Theorem 5 on page 161 of Dummit and Foote's Abstract Algebra, we can write $$G \cong A_1 \times \cdot \cdot \cdot \times A_k$$ where $|A_i|=p_i^{a_i}.$

Question: Is it possible to write $$H \cong B_1\times \cdot \cdot \cdot \times B_k$$ where $|B_i|=p_i^{b_i}$ and $B_i \leq A_i$ for each $i$?

Not always. Counter example:

$$G=\Bbb Z_2\times \Bbb Z_2\;,\;\;\;H=\langle\;(1,1)\;\rangle$$

• How I am reading the OP's notation I don't think this is a counter example. $\vert G \vert = 2^2$ so we would just write $G \cong A_1$ where $\vert A_1 \vert = 4$ (and obviously $A_1$ is the Klein four group) and then $\vert H \vert = 2$ so $H \cong B_1$ with $\vert B_1 \vert = 2$ so $B_1 \leq A_1$.
– user171177
Commented Nov 13, 2014 at 4:06
• Assuming that we are thinking of the primes in the list as being different for different indices
– user171177
Commented Nov 13, 2014 at 4:07
• @Gage I really don't understand what you meant to convey. I gave an example of a group $\;G\;$ , which indeed is Klein's viergruppe, and a subgroup of it which is not the direct product of two subgroups of the factors. Of course $\;B_1\le A_1\;$ , with your notation (which I don't understand why you had to change), but your question was whether any subgroup of a direct product is a direct product of subgroups of the factors, and the above shows this is not always the case. I also don't understand why would we assume the primes in the list are different. Commented Nov 13, 2014 at 4:14
• It isn't my question, I was just saying that how I read the OP's notation that these groups that they say you can break an abelian group up in don't have to be cyclic, just $p$-groups so for example if the group $G = \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3}$ how I read the question as breaking it up is into $A_1 \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ and $A_2 \cong \mathbb{Z}_3 \times \mathbb{Z}_3$ whereas you see it as breaking it up into the 4 cyclic groups. Obviously this is something to clarify with the user asking the question.
– user171177
Commented Nov 13, 2014 at 4:21
• I see your point now, @Gage. Thanks. Yet the OP said nothing so I pressume he meant the most general case. Commented Nov 13, 2014 at 4:29