# if $A$ is Abelian group , $B$ is subgroup of $A$ , Is $B \times A/B \cong A$? [duplicate]

If $A$ is abelian group and $B$ is a subgroup of $A$, $B$ is normal subgroup of $A$.

Is it true that $B \times A/B \cong A$?

I ask because I was watching an online lecture from a course in abstract algebra at Harvard extension school.

And the lecturer (whose name is Peter) was taking about vector spaces and said that if $V$ is a vector space and $W$ is a subspace, then $V/W \times W \cong V$.

So the question which I thought of is: "is this true for all Abelian groups?"

Also, is there a less restricted condition on groups which will make this property hold?

• Try $A=\mathbb Z/4$. – Ted Shifrin Jun 16 '13 at 23:24
• Asked earlier today: math.stackexchange.com/questions/422057/… – Ink Jun 16 '13 at 23:29
• With vector spaces, the underlying field has no nontrivial ideals. But with abelian groups (essentially vector spaces over $\mathbb{Z}$) the underlying ring $\mathbb{Z}$ has nontrivial ideals. So there are nontrivial things to factor out. – alex.jordan Jun 16 '13 at 23:42
• Every vector subspace $W$ is complemented in $V$. Not every subgroup $H$ of an Abelian group $G$ is complemented in (= a factor of) $G$. Typically $\mathbb{Z}_p$ in $\mathbb{Z}_{p^2}$. – Julien Jun 16 '13 at 23:42
• @julien , you means by " $H$ is complemented " is that there exists another subgroup $K$ such that $G = Hk$ ? so the conditions for this property to be true is that $H$ has a complement ? – Fawzy Hegab Jun 16 '13 at 23:45

As suggested by DonAntonio:

We need only take $A =\mathbb Z_4$, $B = \mathbb Z_2$. Then $A/B = \mathbb Z_4/\mathbb Z_2 \cong \mathbb Z_2 = B.\;$ Clearly, $$\mathbb Z_4 \not\cong \mathbb Z_2 \times \mathbb Z_2 = B \times A/B$$

The same "conjecture" would fail for $A = \mathbb Z_9$ and $B = \mathbb Z_3$, for the same reason

Indeed, what we can say is that if $A$ is a finite abelian group, then your conjecture will hold provided there is no prime $p$ such that $p^2$ divides $\,|A|$.

• Gets another TU from me +1 – Amzoti Jun 17 '13 at 0:12
• @amWhy , why does this condition about the non-existence of such $p$ imply that the property hold ? i can guess that this is related with fundamental theorm of finite abelian groups ! but can't recognize the complete connection. – Fawzy Hegab Jun 17 '13 at 0:13
• @MathsLover One variant of the fundamental theorem lets you write groups as products of $p$-power order cyclic groups. If the order of the group is square-free, then there is only one way each particular $p$-power can occur. – Zach L. Jun 17 '13 at 0:14
• Indeed, it relates to the Fundamental Theorem of Finitely Generated groups. Exactly as Zach just summarized. – amWhy Jun 17 '13 at 0:15
• Well, we are talking about the direct product of two groups here, one being a normal subgroup of the abelian group, the other the quotient group. So if the order of a group can be divided by $p^2$, say $|A| = p^2, |B| = p$, then $|A/B| = p$. Now if $A \cong \mathbb Z_{p^2}$, then $A/B \cong B \cong \mathbb Z_p$. And $\mathbb Z_{p^2}\not\cong Z_p \times Z_p$. A similar thing will happen $p^2$ is a factor of the order of abelian A. – amWhy Jun 17 '13 at 0:40

Take $\,A=C_4=$ the cyclic group of order $\,4\,$ , and take $\,B=C_2\,$ , so is

$$C_4\cong C_2\times C_4/C_2=C_2\times C_2\;\;?$$

• no ,$C_4$ can't be isomorphic to $C_2 \times C_2$ using the fundamental theorem for finite abelian groups, so under what conditions do this property hold ? – Fawzy Hegab Jun 16 '13 at 23:31
• @MathsLover For finite abelian groups, you'd get this property if no $p^2$ divided the order of the group. – alex.jordan Jun 16 '13 at 23:44
• @alex.jordan , can you give the reason - proof - behind this result ? – Fawzy Hegab Jun 16 '13 at 23:46