$$D_8=\{(),(1234),(13)(24),(1432),(13),(24),(14)(23),(12)(34) \}$$

Am I right to say that to find the subgroups, we have to make sure that the identity can be generated or is in the group and the inverse of each element can be generated or is in the group?

Is there an easy way to do this though because it seems really time consuming.

  • $\begingroup$ The place to start is by working out the cyclic subgroups and their inclusions. These are by definition the subgroups with a single generator. Then work out subgroups having two or more generators. $\endgroup$ – hardmath May 27 '16 at 17:43
  • $\begingroup$ @hardmath by single generator, do you mean $<a>$ where $a \in D_8$? $\endgroup$ – snowman May 27 '16 at 17:50
  • $\begingroup$ Yes, that's what I mean. $\endgroup$ – hardmath May 27 '16 at 18:46

Think of $D_8$ geometrically - it's the symmetries of a square.

Now, for something to be a "subgroup" of a group, it should be closed under group operations. Now, what does this mean? It means that the subset of $D_8$ that you choose of the group must also form a group.

Now, we can make use of a couple of theorems to figure out the subgroup structure of $D_8$

  1. Lagrange's theorem - the order of a subgroup must divide the order of the entire group. Since $D_8$ has 8 elements, we know that the subgroups can be of order $1$, $2$, $4$ or $8$.

Clearly, the subgroup of order $1$ is the trivial group $\{e\}$, and the subgroup of order $8$ is the entire group $D_8$.

Hence, the subgroups we need to check for are those of order $2$ and $4$. We have a complete classification of the groups of order $2$ and $4$.

We know that the only group of order $2$ is $\mathbb{Z} / 2\mathbb{Z}$.

The groups of order $4$ are the cyclic group $\mathbb{Z} / 4\mathbb{Z}$ and the Klein-$4$ group.

Since we know the different "types" of subgroups we can have, we can now hunt for the subgroups in the dihedral group.

To actually look for the subgroups, use the geometric information of $D_8$. We know that the elements are rotations and a reflection, along with the fact that the rotations form a cyclic subgroup.

Hence, by using both group classification and the "multiplication table" / "geometric information" of $D_8$, we can build up the entire picture.

  • $\begingroup$ Thanks so much but can u give me a quick reference on notation like $\mathbb Z / 4 \mathbb Z $ please. $\endgroup$ – snowman May 27 '16 at 17:44
  • $\begingroup$ $\mathbb{Z}/4\mathbb{Z}$ refers to the cyclic group of order 4. If you've learnt about quotient groups, then you can construct the cyclic group of order $4$ by quotienting the integers($\mathbb{Z}$) with the integers that are multiples of $4$ ($4\mathbb{Z}$) $\endgroup$ – Siddharth Bhat May 27 '16 at 17:56
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    $\begingroup$ While there is only one group of order two up to isomorphism, a specific group like $D_8$ can have more than one subgroup of order two, distinguished by the particular subset of elements chosen. $\endgroup$ – hardmath May 27 '16 at 19:56
  • $\begingroup$ @hardmath - correct, which was why I said "check for subgroup__s__" - that is, check for copies of the same subgroup that show up in different ways :) $\endgroup$ – Siddharth Bhat May 28 '16 at 8:33

Yes, you need to make sure the identity can be generated and the inverse is in the group.

  • $\begingroup$ Actually in a finite group both the identity and inverses are automatically present in any cyclic subgroup. So checking for these separately is not necessary, so long as the cyclic subgroups are included in any larger subsets. $\endgroup$ – hardmath May 27 '16 at 17:38

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