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Find all subgroups of $(\Bbb{Z}_2\times\Bbb{Z}_4,\overline{+})$.

I could find the following subgroups:

$$\begin{array}{ll} H_1=\langle(0,0)\rangle=\{(0,0)\}&\text{(Trivial subgroup)}\\ H_2=\langle(0,1)\rangle=\{(0,1),(0,2),(0,3),(0,0)\}=\langle(0,3)\rangle=H_4\\ H_3=\langle(0,2)\rangle=\{(0,2),(0,0)\}\\ H_5=\langle(1,0)\rangle=\{(1,0),(0,0)\}\\ H_6=\langle(1,1)\rangle=\{(1,1),(0,2),(1,3),(0,0)\}=\langle(1,3)\rangle=H_8\\ H_7=\langle(1,2)\rangle=\{(1,2),(0,0)\}\\ H_9=\Bbb{Z}_2\times\Bbb{Z}_4&\text{(Improper subgroup)} \end{array}$$

That is, $7$ subgroups in total.

I found all the CYCLIC subgroups, but since the group IS NOT CYCLIC, then I need to find the NOT CYCLIC subgroups.

The final answer should be $8$ subgroups; the only subgroup that is NOT CYCLIC is: $$\text{Subgroup that is not cyclic}=\{(0,0),(0,2),(1,0),(1,2)\}.$$ So there are $8$ subgroups in total.

My question is:

How can we find $\{(0,0),(0,2),(1,0),(1,2)\}$? I mean, what should we put inside $\color{red}{\langle(\ldots)\rangle}$?

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    $\begingroup$ if it's not cyclic there's going to be more than one element inside $\langle\rangle$; e.g., $\langle(0,2),(1,0)\rangle$ $\endgroup$ – J. W. Tanner Nov 18 '19 at 17:39
  • $\begingroup$ @J.W.Tanner thank you! Yes I guessed it... But if there's going to be more than one element inside should not we operate them? For example: $\langle(0,0)+(0,1)\rangle=\langle(0,1)\rangle=H_2=H_4$?? $\endgroup$ – manooooh Nov 18 '19 at 17:40
  • $\begingroup$ I don't understand your question in the comment $\endgroup$ – J. W. Tanner Nov 18 '19 at 17:46
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    $\begingroup$ The question is if $\langle x, y \rangle$ is the same as $\langle x+y \rangle$, and the answer is "no". I've updated my answer to address this as well ^_^ $\endgroup$ – HallaSurvivor Nov 18 '19 at 17:47
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    $\begingroup$ True, but my statement broadly stands. There are times where $\langle x + y \rangle = \langle x, y \rangle$, but they are very infrequent. $\endgroup$ – HallaSurvivor Nov 18 '19 at 17:53
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We write $\langle A \rangle$ for a subset (not necessarily subgroup) $A \subseteq G$ to mean the smallest subgroup which contains $A$. So it would be correct in your case to write $$ \{(0,0),(0,2),(1,0),(1,2)\} = \langle \{(0,0),(0,2),(1,0),(1,2)\} \rangle $$

However, this is somewhat unsatisfying, since in the cyclic case we only needed one generator (that is, we only needed one element inside the $\langle - \rangle$). If you are looking for a small set of elements to put inside, you can get away with

$$ \{(0,0),(0,2),(1,0),(1,2)\} = \langle (0,2), (1,0) \rangle.$$

Any subgroup contains $(0,0)$, and any subgroup containing $(1,0)$ and $(0,2)$ must also contain their sum $(1,2)$. Thus {(0,0),(0,2),(1,0),(1,2)} is the smallest subgroup containing $(0,2)$ and $(1,0)$.

Edit: In a comment you ask if we should operate on the elements inside the $\langle - \rangle$. The answer is "no". As you have noticed, if we reduce the elements inside, we can only ever get cyclic subgroups. We allow ourselves multiple elements inside so that we can express more complicated subgroups, such as the example above.

$\langle x + y \rangle$ is always a subgroup of $\langle x, y \rangle$. (Since $x+y \in \langle x, y \rangle$). It is very rarely the case that they are equal, however.


I hope this helps ^_^

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    $\begingroup$ Oh THANK YOU!! So $\langle x,y\rangle$ was the answer!! I have one question: $H_7=\langle (0,2), (1,0) \rangle$ implies that $(0,2)\in H_7$, $(1,0)\in H_7$ and $(0,2)+(1,0)=(1,2)\in H_7$, but it is unclear for me how to operate them to have $(0,0)\in H_7$. $\color{red}{\text{ADD:}}$ Oh, I guess it is the result of operate $(1,2)+(1,2)$, am I right? $\endgroup$ – manooooh Nov 18 '19 at 18:02
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    $\begingroup$ Happy to help ^_^ For the follow up question: $\langle x, y \rangle$ is the smallest subgroup containing $x$ and $y$. But in order to be a subgroup, you have to contain the identity! Similarly, to be a subgroup you have to contain inverses. Because of this we can write $\langle 3 \rangle \leq \mathbb{Z}$ and mean $\{\ldots, -6,-3,0,3,6,\ldots\}$. Even though repeatedly adding $3$ to itself never sums to $0$ (or to any negative numbers!), we know they must still be in $\langle 3 \rangle$, because it is a group. $\endgroup$ – HallaSurvivor Nov 18 '19 at 18:26
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    $\begingroup$ If it helps, you can think of $\langle x, y \rangle$ as all of the elements you can get using $x$, $y$, $x^{-1}$, and $y^{-1}$. Then it becomes clear that we have inverses and the identity ^_^ $\endgroup$ – HallaSurvivor Nov 18 '19 at 18:27
  • $\begingroup$ This is not the case, but when we need to find all subgroups of a group, $\langle x,y\rangle$ can be extended to e.g. three elements: $\langle x,y,z\rangle$ to achieve the same? I am aware that $\langle x,x,x\rangle=\langle x\rangle$ and $\langle x,y,y\rangle=\langle x,y\rangle$ but I want to make sure that if a subgruop can be generated by two or more elements/generators, like $\langle x,y,z\rangle$. $\endgroup$ – manooooh Nov 18 '19 at 18:36
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    $\begingroup$ Of course! $\langle - \rangle$ can take in any subset of $G$ at all. It is sometimes necessary to put infinitely many elements inside. $\endgroup$ – HallaSurvivor Nov 18 '19 at 18:37
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If it's not cyclic there's going to be more than one element inside $\langle\rangle$; e.g., $\langle(0,2),(1,0)\rangle$.

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