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Let $G$ be a group of order $8$. Prove that there is a subgroup of order $4$.

I know that if $G$ is cyclic then there is such a subgroup (if $G=\langle a\rangle$ then the order of $\langle a^2\rangle$ is $4$). But how do I prove this when $G$ is not cyclic? Also, I know that $G$ has an element of order $2$, because the order of $G$ is even. I suspect that assuming all elements of $G$ are of order $2$ somehow leads to a contradiction but am unable to show it. Is this correct or is there a different approach that I'm missing? thanks

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    $\begingroup$ The group might have all elements of order $2$, but then it is abelian (standard exercise), and the result is easy. $\endgroup$ Jul 11 '13 at 13:01
  • $\begingroup$ A more general question is answered here: math.stackexchange.com/questions/306343/…? $\endgroup$ Jul 11 '13 at 13:07
  • $\begingroup$ @GerryMyerson Though that requires quite a bit more than this special case. $\endgroup$ Jul 11 '13 at 13:11
  • $\begingroup$ @Tobias, yes, which is why I'm not suggesting closing this question as a duplicate. But if OP wants to see something a little more hefty, I've pointed to a place to look. $\endgroup$ Jul 11 '13 at 13:12
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You already noted that if $G$ has an element of order $8$ or $4$ then we are done.

Thus we can assume all elements have order $2$ (except the identity element). Then $G$ is abelian (this is a standard exercise, and I am certain it has been asked on MSE several times).

Let $a$ and $b$ be distinct elements of order $2$. Now it is straightforward to check that $\{e,a,b,ab\}$ is a subgroup of order $4$ (where $e$ is the identity element of $G$).

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If $G$ is abelian so according to the Fundamental theorem for finite abelian groups we have: $$G\cong~~~\mathbb Z_8,~~\mathbb Z_4\times\mathbb Z_2,~~\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2$$ So lets assume that $G$ is not abelian. If there is an element in $G$ which is of order $8$ then we have a contradiction. If all non trivial elements of $G$ be of order $2$ so again we have $G=\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2$ which is a new contradiction, so we at least know that there is an elemnt of order $4$. I think we can stop here cause you wanted to know that. For the next you can use the subgroup generated by $x$ to show that $G=\langle y,x\rangle$ in which $y\in G-\langle x\rangle$ and of course $$G\cong \text{Q}_8,~~\text{or}~~\text{D}_8$$

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  • $\begingroup$ @amWhy: Yes Amy. What can I say about the event for my small job here? My English is not good as I can defend my self. :( Some one did the wrong job and someone else should pay the price. I think to myself I have been suggesting everyone I know here to be honest with policy but you see that I am a clueless victim. Sorry for saying these upsetting words but I don't have a tribune for saying it. I have just you to hear my stroy. Sorry Amy. $\endgroup$
    – Mikasa
    Jul 12 '13 at 1:53
  • $\begingroup$ @BabakS.: Sorry that you are going through this! I cannot even understand what happened, how is it possible to lose 5k rep in a day? It just does not make sense and must be a bug of some kind! Hope it gets resolved and it is not likely not something you did my friend! as amWhy said, hang in there! $\endgroup$
    – Amzoti
    Jul 12 '13 at 4:10
  • $\begingroup$ @amWhy: I am warned in a kind way. I don't know what happened because as I read somewhere in Meta, we cannot trace who downvote or upvote others. They think, I have another account and that is why I have more than 5k. Yes, I have and I told rob about it but that account was lost. I have nothing from that account. Even, I wanted rob to merge it for me. I have just 2 questions there. It is ridiculous for me having another active account just making plus. $\endgroup$
    – Mikasa
    Jul 12 '13 at 15:10
  • $\begingroup$ @amWhy:Where will I reach by doing this? More than you or Brian?? Where? What benefit will be for me? ISI Papers? Noble Prize? Believe me, I want to devote all reputations. A real sale is on the way. Sorry my friend. $\endgroup$
    – Mikasa
    Jul 12 '13 at 15:11
  • $\begingroup$ @BabakS. I defended your honor: I told Alexander Gruber that you are the most humble, gracious, and honest person that I know: and I said that with all sincerity. I think we all know you too well: I told him that you have had students who participate, and that I've see you suggest to them (at least, e.g., Flashdesign, that they accept others' answers...that you tend to answer them only when no one else has. I suspect that you've had an admiring student or two who enthusiastically upvoted some of your posts too eagerly, not knowing the consequences. $\endgroup$
    – amWhy
    Jul 12 '13 at 15:16
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This is a direct consequence of the following theorem.

Let $G$ be a group of order $p^{n}$ for $p$ a prime. Then for each $m$ with $0 \leq m \leq n$, $G$ has a subgroup of order $p^{m}$.

The proof is here.

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    $\begingroup$ This is the result pointed to by Gerry Meyerson. The proof takes considerably more than is needed for this special case. $\endgroup$ Jul 11 '13 at 13:24
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I would consider elements of different orders and take cases accordingly.

You already are happy with what to do if there is an element of order 8 (and hence the group is generated by this element and cyclic).

Next, if there is a generator of order four, there clearly can only be one other generator, and this must be of order two. If we have $a$ and $b$ of orders 2 and 4 respectively you can easily check check what all the elements are, and it should be obvious what the subgroup of order 4 is.

Lastly, if you have a generator of order two, you could have another generator of order 4, but we have already considered this. The only other alternative is to have two more generators of order 2, which must commute (else the group they generate would be too large) and again, if you consider the abelian group generated by $a$, $b$ and $c$, all of order two, it should be clear how you can generate a group of order four.

Note that this is not the most elegant solution, but it is (probably) the most concrete. And please ask if anything is not clear.

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  • $\begingroup$ "if there is an element of order four, there clearly can only be one other element" Huh? Surely, that's not what you meant to write. $\endgroup$ Jul 11 '13 at 13:13
  • $\begingroup$ If there is an element of order $4$, you are clearly done. And the fact that the elements of order $2$ will commute follows only once you use that all the elements have order $2$. $\endgroup$ Jul 11 '13 at 13:13
  • $\begingroup$ @GerryMyerson sorry, I meant to talk about generators, not elements. Thank you! $\endgroup$
    – Joe Tait
    Jul 11 '13 at 13:15
  • $\begingroup$ A group of order $8$ cannot have a generator of order $4$. $\endgroup$ Jul 11 '13 at 13:17
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    $\begingroup$ When you say G is generated by a, b and c you mean G = {e, a, b, c, ab, ac, b*c...}? $\endgroup$ Jul 11 '13 at 13:36

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