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Here is the exact sequence of abelian groups I am studying:

$$0 \to \mathbb Z/2 \to B \to \mathbb Z/2 \to 0 $$

Can I say that $B \cong \mathbb Z/2$ or $B \cong \mathbb Z/2 \oplus \mathbb Z/2$? Is $B \cong \mathbb Z/4$ a possibility also, if so, why? If not, why also?

Could someone clarify this to me please?

Edit:

What if I have the following sequence instead $$0 \to \mathbb Z/3 \to B \to \mathbb Z/5 \to 0 $$

How should my way of thinking differ?

Intuition
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Emptymind
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  • No, $B\cong Z/2$ is not one 9f the options. – Thomas Andrews May 03 '24 at 16:22
  • why not? @ThomasAndrews – Emptymind May 03 '24 at 16:23
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    In general, if $A,C$ are finite and $0\to A\to B\to C\to 0$ is exact, then $|B|=|A|\cdot|C|$ Here, if $B\cong Z/2,$ then $Z/2\to B$ has to be $1$, so the function has to be the identity, and if $B\to Z/2$ has to be onto, so that map also has to be the identity. But then $Z/2\to B\to Z/2$ can't be exact, because the kernel of the right map is ${0}$ and the image of the left map is all of $B.$ – Thomas Andrews May 03 '24 at 16:27
  • I got it thank you very much! Do you know where is the idea of the order you mentioned pointed out in Dummit and Foote? – Emptymind May 03 '24 at 16:30
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    Sorry, I don't have that book, but the basic idea is elementary. $g:B\to C\to 0$ exact means this function is onto, so by a standard theorem, $B/\ker g=C.$ Then $A\to_f B\to_gC$ exact means $\operatorname{im} f=\ker g.$ So $B/\operatorname{im} f\cong C.$ But $0\to A\to_f B$ exact means $f$ is one-to-one, so $|\operatorname{im}f|=|A|.$ If $M/N=P$ in abelian groups then $|M|=|N|\cdot |P|.$ (This is actually true even for infinite cases - it becomes a product of infinite cardinals.) – Thomas Andrews May 03 '24 at 16:40
  • $B=A\oplus C$ is always (and obviously) a possibility in $0\to A\to B\to C\to0$. The exact sequence $0\to\Bbb Z_2\to\Bbb Z_4\to\Bbb Z_2\to0$ is given by the surjective homomorphism $\Bbb Z_4\to\Bbb Z_2,[k\bmod4]\mapsto[k\bmod2]$. – Anne Bauval May 03 '24 at 16:43
  • Related. Hint for your final question: there is a unique group of order $15$. – Anne Bauval May 03 '24 at 16:57
  • what if it was 7 not five? @AnneBauval – Emptymind May 03 '24 at 17:39
  • https://meta.stackexchange.com/questions/43478/exit-strategies-for-chameleon-questions https://groupprops.subwiki.org/wiki/Groups_of_order_21 – Anne Bauval May 03 '24 at 17:44
  • I did not get the idea, can you explain it simply please?@AnneBauval – Emptymind May 03 '24 at 17:52
  • @AnneBauval why you spoke only about the surjective homomorphism in your comment before last? – Emptymind May 03 '24 at 18:24
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    Because it is sufficient to determine completely the exact sequence. $A$ is the kernel of the surjective morphism $B\to C$. – Anne Bauval May 03 '24 at 18:29
  • but not all $\mathbb Z_4$ go to zero @AnneBauval by the map you gave. – Emptymind May 03 '24 at 18:37
  • Of course! The kernel $A$ of this homomorphism is (happily) not all $\Bbb Z_4$. It is of order $|B|/|C|=4/2=2$. – Anne Bauval May 03 '24 at 18:40
  • so this is the map in case $B \cong Z/2 \oplus Z/2,$ not in case of $B \cong Z/4.$ Am I correct? @AnneBauval – Emptymind May 03 '24 at 18:46
  • No, come on! The map we are talking about is the one about which you asked me, 2 hours ago: "why you spoke only about the surjective homomorphism in your comment before last?", i.e., as I wrote in the said comment: "$\Bbb Z_4\to\Bbb Z_2,[k\bmod4]\mapsto[k\bmod2]$". – Anne Bauval May 03 '24 at 21:19

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Remember that saying that $ A\stackrel{f}{\to} B\stackrel{g}{\to} C$ "is exact at $B$" means that $\mathrm{im}(f)=\ker(g)$. To say that a sequence $$\cdots \stackrel{f_{i-1}}{\longrightarrow} A_{i-1}\stackrel{f_i}{\longrightarrow}A_i\stackrel{f_{i+1}}{\longrightarrow}A_{i+1}\stackrel{f_{i+2}}{\longrightarrow}\cdots$$ "is exact" means that it is exact at each $A_i$. To say that you have a "short exact sequence" $$0\to A\stackrel{f}{\longrightarrow} B \stackrel{g}{\longrightarrow} C\to 0$$ means that this is exact at $A$, $B$, and $C$. In particular:

  1. $\mathrm{Im}(f)=\ker(g)$;
  2. The kernel of $f$ is the image of the map $0\to A$... which is $\{0\}$. So $f$ is one-to-one.
  3. The image of $g$ is the kernel of the map $C\to 0$... which is all of $C$, so $g$ is surjective.

That means that if you say that you have a short exact sequence $$0 \to A \to B \to C \to 0,$$ means that $B$ has a subgroup $A'$ isomorphic to $A$, and that the quotient $B/A'$ is isomorphic to $C$.

From $$0 \to \mathbb{Z}/2\mathbb{Z}\to B\to\mathbb{Z}/2\mathbb{Z}\to 0$$ we know that $B$ has a subgroup $A$ that is isomorphic to $\mathbb{Z}/2\mathbb{Z}$, and that $B/A$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$. This tells you that $|A|=2$, and that $|B/A|=2$, and therefore that $|B|=4$. In particular, it certainly cannot be the case that $B=\mathbb{Z}/2\mathbb{Z}$.

Given the tags in the post, I am working in the context of abelian groups only, so... which abelian groups of order $4$ have a subgroup of order $2$ with quotient of order $2$? Well... all of them. So $B$ can be either $\mathbb{Z}_4$, or $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$.

If you have $$0 \to \mathbb{Z}/3\mathbb{Z} \to B \to \mathbb{Z}/5\mathbb{Z}\to 0$$ then you have an abelian group $B$ with a subgroup $A$ of order $3$, and such that $B/A$ is cyclic of order $5$. Hence $5 = |B/A| = |B|/|A| = |B|/3$, so $|B|=15$. There is one and only one abelian group of order $15$ (in fact, there is one and only one group of order $15$), so $B$ is that group: $\mathbb{Z}/15\mathbb{Z}$. The same thing happens with $$0 \to \mathbb{Z}/3\mathbb{Z}\to B\to\mathbb{Z}/7\mathbb{Z}\to 0$$ which you ask in the comments: $B$ must be the cyclic group of order $21$. More generally if $A$ and $C$ are abelian groups of relatively prime order, then the exact sequence $$0\to A\to B\to C\to 0$$ forces $B$ to be isomorphic to $A\oplus C$. In the two latter examples above, $\mathbb{Z}/3\mathbb{Z}\oplus\mathbb{Z}/5\mathbb{Z}$ (which is isomorphic to $\mathbb{Z}/15\mathbb{Z}$) and $\mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/7\mathbb{Z}$ (which is isomorphic to $\mathbb{Z}/21\mathbb{Z}$).

Arturo Magidin
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