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If $ H $ is a normal subgroup of $ G $, is $ G/H \times H \cong G $?

For example, I think $ \mathbb{Z}/2 \mathbb{Z} \times 2 \mathbb{Z} \cong \mathbb{Z} $. I would construct a map as follows: \begin{align} \phi: \mathbb{Z}/2 \mathbb{Z} \times 2 \mathbb{Z} &\longrightarrow \mathbb{Z}; \\ (a,b) &\longmapsto a + b. \end{align} If it is not true in general, then are there a few criteria to show when it is true?

Thanks!

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    $\begingroup$ For the record, I would like math to work this way. $\endgroup$ – Ziggy Mar 5 '13 at 6:45
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    $\begingroup$ Sadly math does not work this way. Notice that your LHS is not cyclic and has elements of finite order. $\endgroup$ – JSchlather Mar 5 '13 at 6:48
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    $\begingroup$ See en.wikipedia.org/wiki/Semidirect_product and en.wikipedia.org/wiki/Group_extension . This is almost never true. $\endgroup$ – Qiaochu Yuan Mar 5 '13 at 6:51
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    $\begingroup$ Hmm... it seems so strange and sad that we can 'kill' H in G, but we can't fix the damage we've done? Is there maybe a different operation for 'resuscitating' H in G? I'm just concerned for H is all. $\endgroup$ – Ziggy Mar 5 '13 at 7:10
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    $\begingroup$ It works for vector spaces. Go vector spaces, go! $\endgroup$ – user1729 Mar 5 '13 at 10:01
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No. If $A\times B\cong G$ then both $A$ and $B$ (respectively their images under the isomorphism) are normal in $G$. Your example with $\mathbb Z/2\mathbb Z$ does not work because $\mathbb Z/2\mathbb Z$ is not normal in $\mathbb Z$, yes it is not even a subgroup (no nonzero element $x\in \mathbb Z$ has the property $x+x=0$).

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Let me add a couple of comments to the other excellent answers.

In some cases there is a subgroup $K$ of $G$ such that $G = H K$ and $H \cap K = 1$. Then $G/H \cong K$, and $G$ is a semidirect product (a.k.a. split extension) $H \rtimes K$ of $H$ by $K$. The semidirect product can be regarded as a generalization of the direct product, in which only one of the factors is normal.

In general, even such a subgroup $K$ of $G$ will not exist. Define then $K = G/H$. Now $G$ is said to be an extension of $H$ by $K$. Things get immediately very difficult here. However, one can define a section as a map $\sigma : K \to G$ such that $\varphi \circ \sigma = \textbf{1}_{K}$. (Here $\varphi : G \to K = G/H$ is the canonical map.) $\sigma(K)$ will not be a subgroup of $G$ (unless $G$ is a split extension of $H$ by $K$ as above), but the multiplication between elements of $\sigma(K)$ will have to be corrected by an element of $H$. This yields what is called a $1$-cocycle, and from now on you are in the territory of group cohomology.

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  • $\begingroup$ Is it necessary for the map $\sigma$ to be a homomorphism? Or is it just any well-defined map that works as a right-hand inverse for the canonical projection? $\endgroup$ – Miguel Angel Alarcon Bustos Nov 6 '17 at 21:43
  • $\begingroup$ @MiguelAngelAlarconBustos, apologies for the late answer. There is a $\sigma$ which is a homomorphism if and only if $G$ is a split extension. Think of the case when $G$ is cyclic of order $4$, and $K$ is the subgroup of order $2$. $\endgroup$ – Andreas Caranti Nov 17 '17 at 12:45
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Any cyclic group whose order is not squarefree contains at least one normal subgroup for which this fails, as $C_{p^2m}\not\cong C_{p}\times C_{pm}$.

You may be looking for the Schur-Zassenhaus theorem.

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Two trivial cases where this is true are $H = \{ 1 \}$ and $H = G$. Also, there exist many groups for which these are the only cases. One example is $S_3$, the nonabelian group of order $6$. If $S_3 \cong A \times B$, then $A = \{1\}$ or $A = S_3$.

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  • $\begingroup$ $\mathbb Z_4$ is an even smaller example! $\endgroup$ – yatima2975 Mar 5 '13 at 9:02
  • $\begingroup$ @yatima2975: you're right, Thanks. The cyclic groups of orders $1$, $2$, and $3$ are also examples, and more generally cyclic groups of prime power order give examples. $\endgroup$ – spin Mar 5 '13 at 17:28
  • $\begingroup$ Orders 1,2 and 3 don't work because there a subgroup is either trivial or the whole group! $\endgroup$ – yatima2975 Mar 5 '13 at 17:39
  • $\begingroup$ I meant to give examples of indecomposable groups, not necessarily those where the statement fails. It fails for an indecomposable group unless it is of prime order or trivial $\endgroup$ – spin Mar 5 '13 at 20:51
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It's worth noting that even when $G \cong H \times K$, in order for $G/H \times H$ to be isomorphic to $G$, you need to pick the "right" copy of $H$.

For example, let $G = C_2 \times C_4$, let $H = C_2$, and let $F = \{0\} \times \{0, 2\}$ be a subgroup of $G$. Then $F \cong H$ and, since $G$ is abelian, $F$ is a normal subgroup. However, $$(G/F) \times H \cong C_2 \times C_2 \times C_2 \ncong G.$$

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