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This is Exercise 8.28 of Gallian's "Contemporary Abstract Algebra".

Answers that use only methods from the textbook prior to the exercise are preferred.

Here $G_1\oplus G_2$ is the external direct product of $G_1$ by $G_2$.

Here $G_1\le G_2$ means $G_1$ is a subgroup or equal to the group $G_2$.

The Question:

Find a subgroup of $\Bbb Z_4\oplus\Bbb Z_2$ not of the form $H\oplus K$ for some $H\le \Bbb Z_4, K\le \Bbb Z_2$.

Thoughts:

I must confess: I cheated here a little bit by looking up the subgroups of $\Bbb Z_4\times \Bbb Z_2$. But notice the difference in notation! I think in terms of just plain old direct products (because aren't internal and external direct products equivalent? Yes! But this is not established in the textbook yet; indeed, the former is not even mentioned at this point).

It appears to me to be a trick question. Here are the subgroups of $\Bbb Z_4\times \Bbb Z_2$. Where is the subgroup of the desired form?

My guess is that there's some technical aspect of external direct products that is being emphasised here.

Please help :)

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  • $\begingroup$ What does "internal direct product" mean? Are you assuming $\mathbb Z_4,\mathbb Z_2$ are subgroups of some larger group? $\endgroup$ – Wojowu Mar 3 at 15:29
  • $\begingroup$ Sorry, @Wojowu: I got them the wrong way around! Hang on a second while I edit the question . . . $\endgroup$ – Shaun Mar 3 at 15:31
  • $\begingroup$ Also, the symmetric groups by Cayley's theorem, I would imagine, @Wojowu. $\endgroup$ – Shaun Mar 3 at 15:33
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    $\begingroup$ Internal direct products are mentioned in Section 9 of the textbook, @Wojowu; I'm on Section 8. $\endgroup$ – Shaun Mar 3 at 15:37
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    $\begingroup$ @Shaun In a small case like this, "shut up and calculate" might be my preferred approach. But there are some theoretical considerations which may help it go faster. For instance, if a subgroup is of the form $H\oplus K$ and contains $(a,b)$, then it must also contain $(a,0)$ and $(0,b)$. So taking an element of the form $(a,b)$ with $a,b\neq0$ and looking at the cyclic group generated by that element would be my first attempt. $\endgroup$ – Arthur Mar 3 at 16:02
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The cyclic group generated by $(1,1)$ fits. It maps surjectively under both projections, but it is not the whole group, so you cannot write it as the direct product of two subgroups.

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