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I need to determine the subgroups of the dihedral group of order 4, $D_4$.

I know that the elements of $D_4$ are $\{1,r,r^2,r^3, s,rs,r^2s,r^3s\}$

But I don't understand how to get the subgroups..

Dietrich Burde
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amir
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  • The elements don't do you any good unless you know generators/relations, or something else about D4 to get you the group structure. What do you know about D4 apart from the names of the elements? – Tyler Oct 19 '13 at 16:19
  • @Tyler we know that $D_4$ is generated by the rotation $r$ and the reflection $s$ – amir Oct 19 '13 at 16:20
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    It is very disrespectful to delete a question once you get an answer. You shouldn't do that again. – Pedro Oct 19 '13 at 17:02
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    @Pedro: in the OP's defense, the question was deleted only 39 seconds after the answer was posted. It is possible the the OP hadn't seen the answer appear yet. – hmakholm left over Monica Oct 19 '13 at 17:05
  • @HenningMakholm Noted. – Pedro Oct 19 '13 at 17:16
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    @PedroTamaroff It is true I didn't see it.. – amir Oct 19 '13 at 17:23

2 Answers2

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By Lagrange's Theorem, the possible orders are $1, 2, 4,$ and $8$.

The only subgroup of order $1$ is $\{1\}$ and the only subgroup of order $8$ is $D_4$.

If $D_4$ has an order $2$ subgroup, it must be isomorphic to $\mathbb{Z}_2$ (this is the only group of order $2$ up to isomorphism). Such a group is cyclic, it is generated by an element of order $2$. Are there any such elements in $D_4$?

If $D_4$ has an order $4$ subgroup, it must be isomorphic to either $\mathbb{Z}_4$ or $\mathbb{Z}_2\times\mathbb{Z}_2$ (these are the only groups of order $4$ up to isomorphism). In the former case, the group is cyclic, it is generated by an element of order $4$. Are there any such elements in $D_4$? In the latter case, the group is generated by two commuting elements of order $2$. Are there any such pairs of elements in $D_4$?

In summary, first find all the elements of order $2$ and all the elements of order $4$; each of them generates a cyclic subgroup. Then consider pairs of elements of order $2$ to find which of them generate subgroups isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$.

Michael Albanese
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  • This is a very stupid question but still I want to ask: how to prove that any group of order $4$ is isomorphic to $\mathbb {Z}_4$ or $\mathbb {Z}_2×\mathbb {Z}_2$? Also, how do u conclude that "In the latter case, the group is generated by two commuting elements of order $2$" in ur answer? – Arthur Dec 26 '22 at 10:17
  • @Franklin: Either $G$ has an element of order four, in which case it is isomorphic to $\mathbb{Z}_4$, or it doesn't. In the latter case, every non-identity element has order two by Lagrange's Theorem; let $a, b \in G$ be two distinct such elements. Since $G$ has order four, we see that $ab$ and $ba$ must be equal, so $G \cong\langle a\rangle\times\langle b\rangle \cong \mathbb{Z}_2\times\mathbb{Z}_2$. The two commuting elements which generate $G$ are $a$ and $b$. – Michael Albanese Dec 26 '22 at 15:03
  • Thank you so much for responding ! But I dont get how do u conclude that $G\cong \langle a\rangle ×\langle b\rangle $ ?... – Arthur Dec 26 '22 at 15:59
  • @Franklin: $G$ has four elements: $1$, $a$, $b$, and $ab$. There is an isomorphism $\langle a\rangle\times\langle b\rangle \to G$ given by $(1, 1) \mapsto 1$, $(a, 1) \mapsto a$, $(1, b) \mapsto b$, and $(a, b) \mapsto ab$. – Michael Albanese Dec 26 '22 at 16:01
  • Thank you so much ! I remain much grateful.... – Arthur Dec 26 '22 at 16:31
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The notes by K. Conrad have a nice answer: the dihedral group $D_n$ is generated by a rotation $r$ and a reflection $s$ subject to the relations $r^n=s^2=1$ and $(rs)^2=1$.

Proposition: Every subgroup of $D_n$ is cyclic or dihedral. A complete listing of the subgroups (including $1$ and $D_n$) is as follows:

$(1)$ $\langle r^d \rangle$ for all divisors $d\mid n$.
$(2)$ $\langle r^d,r^is \rangle$, where $d\mid n$ and $0\le i\le d-1 $.

Very nice pictures of the subgroup diagram of $D_4$ can be found here.

Dietrich Burde
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    Thank you very much, I was searching the subgroups of the dihedral group. I would love if your answer was to the question: What are the subgroups of ANY dihedral group? If it was like that, I would have found it in google easier! – Santropedro Jan 16 '17 at 00:59
  • Oh, you interpreted my comment not like I intended! Indeed your answer is for all the dihedral groups. I was trying to say that it would be better if the title of this question was for any dihedral group, making the question appear in my searches! I was avoiding questions that only worked for some specific dihedral group. Then I tried with this question and your answer helped me. – Santropedro Jan 16 '17 at 14:17
  • @Santropedro Oh, I apologize! It sounded liked you would love if my answer would have been for ANY dihedral group - reading it again it is different of course. – Dietrich Burde Jan 16 '17 at 15:10
  • Yes, communicating is hard sometimes! Note: Your answer helped me a lot, thanks! – Santropedro Jan 16 '17 at 17:27