# Thread: A Few Questions Involving Subgroups

1. ## A Few Questions Involving Subgroups

1) Let G = <x> be a cyclic group of order n. Show that x^m is a generator of G if and only if (m,n) = 1. Thus the number of generators of a cyclic group of order n is the number of integers m in the set (0,1,...,n-1) such that (m,n) = 1. This is called Euler's function and plays a prominent role in number theory.

2) Let mZ and nZ be subgroups of (Z, +). What condition on m and n is equivalent to mZ is a subset of nZ? What condition on m and n is equivalent to mZ union nZ being a subgroup of (Z,+)?
(I don't really even get what this is asking...)

3) Prove that the intersection of two subgroups of a group G is itself a subgroup of G.

4) Give an example of a group G and a subset H of G such that H is closed under multiplication but H is not a subgroup of G.

5) a) Show that it is impossible for a group G to be the union of two proper subgroups.
b) Give an example of a group that is the union of three proper subgroups.

2. Originally Posted by Janu42
1) Let G = <x> be a cyclic group of order n. Show that x^m is a generator of G if and only if (m,n) = 1. Thus the number of generators of a cyclic group of order n is the number of integers m in the set (0,1,...,n-1) such that (m,n) = 1. This is called Euler's function and plays a prominent role in number theory.

2) Let mZ and nZ be subgroups of (Z, +). What condition on m and n is equivalent to mZ is a subset of nZ? What condition on m and n is equivalent to mZ union nZ being a subgroup of (Z,+)?
(I don't really even get what this is asking...)

3) Prove that the intersection of two subgroups of a group G is itself a subgroup of G.

4) Give an example of a group G and a subset H of G such that H is closed under multiplication but H is not a subgroup of G.

5) a) Show that it is impossible for a group G to be the union of two proper subgroups.
b) Give an example of a group that is the union of three proper subgroups.
3) here

3. Originally Posted by Janu42
1) Let G = <x> be a cyclic group of order n. Show that x^m is a generator of G if and only if (m,n) = 1. Thus the number of generators of a cyclic group of order n is the number of integers m in the set (0,1,...,n-1) such that (m,n) = 1. This is called Euler's function and plays a prominent role in number theory.
To show $\left< x^m \right> = G$ it is sufficient to show the order of $x^m$ is $n$. Say that $k$ is the order of $x^m$ then it means $x^{km} = e$. But $x$ has order $n$ therefore $n|km \implies n|k \implies n=k$ since $(n,m)=1$. Conversely, say that $\left< x^m \right> = G$ and say that $(n,m)=d>1$. We will arrive at a contradiction. Note that $(x^m)^{n/d} = (x^n)^{m/d} = e$. But then it means the order of $x^m$ is less than $n$ since $n/d - this is a contradiction.

2) Let mZ and nZ be subgroups of (Z, +). What condition on m and n is equivalent to mZ is a subset of nZ? What condition on m and n is equivalent to mZ union nZ being a subgroup of (Z,+)?
(I don't really even get what this is asking...)
This problem is asking if $m\mathbb{Z}\subseteq n\mathbb{Z}$ what can you conclude about the relationship between $n\text{ and }m$.

4) Give an example of a group G and a subset H of G such that H is closed under multiplication but H is not a subgroup of G.
Let $G = \mathbb{R}$ and let $H = \{ 2^r | r>0\}$.

5) a) Show that it is impossible for a group G to be the union of two proper subgroups.
Hint: If a union of two subgroups is a group then one subgroup is contained in another one.

b) Give an example of a group that is the union of three proper subgroups.
Let $G=\mathbb{Z}_2\times \mathbb{Z}_2$.
Can you think of it now?

4. For 5(a), why does one have to be contained in the other? Like, if G = {1,2,3,4,5,6}, couldn't one subgroup be {1,2,3} and the other {4,5,6}?

For (b), would it be (0,0), (0,1), and (1,0) as the proper subgroups?

Also for 4, this is closed under multiplication obviously. But it is not a subgroup because it is not closed under inverses?

5. And 2 more:

1) Let G be a group and let H be a nonempty subset of G such that whenever x,y are in H we have xy^-1 in H. Prove that H is a subgroup of G.

2) Let G be a group and let a be some fixed element of G. Let H be a subgroup of G and let aHa^-1 be the subset of G consisting of all elements that are of the form aha^-1, with h in H. Show that aHa^-1 is a subgroup of G. It is called the conjugate subgroup of H by a.

6. Originally Posted by Janu42

Also for 4, this is closed under multiplication obviously. But it is not a subgroup because it is not closed under inverses?
there is no identity element

7. Oh, for 5, does something have to do with the fact that the identity is in every subgroup?

8. Originally Posted by Janu42
Oh, for 5, does something have to do with the fact that the identity is in every subgroup?
By definition a subgroup needs to have an identity element.

9. And for 5(b), I don't know if what I said is right. Is it the union of (0,0), (0,1), and (1,0)?

10. Originally Posted by Janu42
And for 5(b), I don't know if what I said is right. Is it the union of (0,0), (0,1), and (1,0)?
Let $H_1 = \{ (0,0),(1,0)\}, H_2 = \{ (0,0),(0,1)\},H_3 = \{ (0,0),(1,1)\}$.
Then it means $H_1\cup H_2\cup H_3 = \mathbb{Z}_2\times \mathbb{Z}_2$.

11. Ahh, right. OK. I had the right idea but overlooked the fact that the identity needs to be in every subgroup. Thanks.

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# example of a group that is the union of three proper subgroups

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