# How do you find the normal subgroups of DN?

## How do you find the normal subgroups of DN?

In Dn, every subgroup of 〈r〉 is a normal subgroup of Dn; these are the subgroups 〈rd〉 for d | n and have index 2d. This describes all proper normal subgroups of Dn when n is odd, and the only additional proper normal subgroups when n is even are 〈r2,s〉 and 〈r2, rs〉 with index 2.

## How do you find the normal subgroups of D4?

(a) The proper normal subgroups of D4 = {e, r, r2,r3, s, rs, r2s, r3s} are {e, r, r2,r3}, {e, r2, s, r2s}, {e, r2, rs, r3s}, and {e, r2}. To see this note that s is conjugate to r2s (conjugate by r), so if a subgroup contains s for it to be normal it must contain r2s.

**How many normal subgroups does D8 have?**

4 subgroups

The lattice of subgroups of D8 is given on [p69, Dummit & Foote]. All order 4 subgroups and 〈r2〉 are normal. Thus all quotient groups of D8 over order 4 normal subgroups are isomorphic to Z2 and D8/〈r2〉 = {1{1,r2},r{1,r2},s{1,r2}, rs{1,r2}} ≃ D4 ≃ V4.

**What is normal subgroup with example?**

A normal subgroup is a subgroup that is invariant under conjugation by any element of the original group: H is normal if and only if g H g − 1 = H gHg^{-1} = H gHg−1=H for any. g \in G. g∈G. Equivalently, a subgroup H of G is normal if and only if g H = H g gH = Hg gH=Hg for any g ∈ G g \in G g∈G.

### What is an example of a normal subgroup of Dn?

Solution Trivial examples of normal subgroups are GL(2,R) itself, and {eGL(2,R)} = {I}. A less trivial example is SL(2,R), which is the kernel of the determinant and so a normal subgroup (since the determinant is a homomorphism).

### How many subgroups of order n does DN have?

D(24) has 64 subgroups….

n | Number of Subgroups of D(n) |
---|---|

3 | 6 |

4 | 10 |

5 | 8 |

6 | 16 |

**How many subgroups are in a normal D4?**

Thus, D4 have one 2-element normal subgroup and three 4-element subgroups.

**How do you find the normal subgroups of D8?**

The lattice of subgroups of D8 is given on . All order 4 subgroups and 〈r2〉 are normal. Thus all quotient groups of D8 over order 4 normal subgroups are isomorphic to Z2 and D8/〈r2〉 = {1{1,r2},r{1,r2},s{1,r2}, rs{1,r2}} ≃ D4 ≃ V4.

#### What are the subgroups of D12?

(d) In D12, we see that there are three normal subgroups of index 2, namely C6 and two D6s. Moreover, C6 has two composition series C6 >C2 > {1} and C6 > C3 > {1}, while D6 has only one, namely D6 >C3 > {1}. So there are four composition series for D12. 2 The normal subgroups of S4 are A4, V4 (the Klein group) and {1}.

#### What is the unique subgroup of D6 having order 3?

H = { i d, R 2, R 4 } is the unique subgroup of D 6 having order 3. It is a normal subgroup, because we have R i ( R 2) R − 1 = R 2, R i ( R 4) R − 1 = R 4, and F 1 R 2 F 1 − 1 = R 4, F 2 R 2 F 2 − 2 = R 4 and so on.

**Is the quotient group d6/h a normal group?**

It is a normal subgroup, because we have R i ( R 2) R − 1 = R 2, R i ( R 4) R − 1 = R 4, and F 1 R 2 F 1 − 1 = R 4, F 2 R 2 F 2 − 2 = R 4 and so on. We just can check this from the Cayley table, that g H g − 1 ⊆ H for all g ∈ D 6. It follows that the quotient group D 6 / H has order 4.

**How to find a normal subgroup of order $4$?**

One possibility is the trivial group, order $1$. We cannot get a normal subgroup of orders $2$ or $3$ (in particular, you $H$ cannot possibly be normal). The only way to get a subgroup of order $4$ is to take the class of the identity and the class of the product of two transpositions.

## What are the 8 elements of D4?

D4 has 8 elements: 1,r,r2,r3, d. 1,d2,b1,b2, where r is the rotation on 90◦, d. 1,d2 are ﬂips about diagonals, b1,b2 are ﬂips about the lines joining the centersof opposite sides of a square. Let N be a normal subgroup of D4.