Does cyclic group have nontrivial subgroups?

If a nontrivial group has no proper nontrivial subgroup, then it is a cyclic group of prime order. In other words, it is generated by a single element whose order is a prime number.

What is a proper nontrivial subgroup?

As you know the identity element is trivial subgroup, all other subgroups are nontrivial and G is the improper subgroup of G, and all others are proper subgroups. Now a proper non trivial subgroup means it is neither identity nor G it. This is subgroup other than Identity and G itself.

How do you find non trivial subgroups?

By Lagrange, every subgroup of U(73) has order d∣ϕ(73)=72 Consider the subgroup U=⟨2⟩, generated by 2. Since 29≡1mod73, U is a non-trivial subgroup of U(73), having order d=9.

What is a nontrivial cyclic group?

Let G be a group whose identity element is e. If g has infinite order, then ⟨g⟩ is an infinite cyclic group. If |g|=n, then ⟨g⟩ is a cyclic group with n elements. Thus, every group which is non-trivial has at least one cyclic subgroup which is also non-trivial.

Can a cyclic group have only one generator?

Thus a cyclic group may have more than one generator. However, not all elements of G need be generators. For example 〈−1〉 = {1,−1} = G so −1 is not a generator of G. 7 = the group of units of the ring Z7 is a cyclic group with generator 3.

What is a proper subgroup of a group?

A proper subgroup of a group G is a subgroup H which is a proper subset of G (that is, H ≠ G). This is usually represented notationally by H < G, read as “H is a proper subgroup of G”. If H is a subgroup of G, then G is sometimes called an overgroup of H.

How many non-trivial proper subgroup does the group z8 have?

Center (group theory)

Is not necessarily a property of a group?

__________ is not necessarily a property of a Group. Explanation: Grupoid has closure property; semigroup has closure and associative; monoid has closure, associative and identity property; group has closure, associative, identity and inverse; the abelian group has group property and commutative.

How do you find the cyclic subgroups of the UN?

Let U(n) be a cyclic group. The order of every subgroup of U(n) is a divisor of |U(n)| and conversely, for every divisor d of |U(n)|, there is a unique cyclic subgroup of order d. Hence, the number of distinct subgroups of U(n) is the number of divisors of |U(n)|.

How to prove that $G$ is finite of prime order?

(2 answers) Closed 3 years ago. We know the following fact from gorup theory: If $G$ is a group of prime order then it has no nontrivial subgroups. Lets try to prove the converse statement: If $G$ has no nontrivial subgroups, show that $G$ must be finite of prime order.

Is a group with no proper nontrivial subgroups cyclic?

There is a lemma that says if a group $G$ has no proper nontrivial subgroups, then $G$ is cyclic. And here is the proof of the lemma: Suppose $G$ has no proper nontrivial subgroups.

Is $G$ A cyclic group?

But $G$ has no proper subgroups, so it must be that $\\langle a angle = G$. Thus $G$ is cyclic, by definition of a cyclic group. But here i do not understand the following: Why must $\\langle a angle$ be a subgroup of $G$?

How do you prove that a group is cyclic?

Let us suppose that G is a cyclic group generated by a i.e. G = {a}. If another group H is equal to G or H = {a}, then obviously H is cyclic. So let H be a proper subgroup of G. Therefore, the elements of H will be the integral powers of a. Therefore, H contains elements that are positive as well as negative integral powers of a.