# Thread: Finite Abelian Group

1. ## Finite Abelian Group

Question: G is a finite abelian group of order n. Let d|n.
Prove G has a subgroup of order 'd'.

I know that if p^a (where p is a prime and a any +ve integer) divides n, then G has a subgroup of order p^a

I was trying to use this for the question. Is this correct approach? I have not read the generic theory of abelian groups, so would want to avoid that as much.

Any help/pointers please?

Thanks

2. Originally Posted by aman_cc
Question: G is a finite abelian group of order n. Let d|n.
Prove G has a subgroup of order 'd'.

I know that if p^a (where p is a prime and a any +ve integer) divides n, then G has a subgroup of order p^a

I was trying to use this for the question. Is this correct approach? I have not read the generic theory of abelian groups, so would want to avoid that as much.

Any help/pointers please?

Thanks

Of course this is enough, but you must know:

i) If H,K are two sbgps. of a group G, then HK is again a sbgp. iff HK = KH

ii) If N is a NORMAL sbgp. of a group G, then NK is always a sbgp. of G for any sbgp. K of G

iii) For any two sbgps. of an ABELIAN group G, HK is again a sbgp. of G.

iv) For H,K sbgps. of G, |HK| = |H||K|/|H /\ K|

Take some minutes to fully understand the above and then try to put things together.
Note also that (i ==> (ii) ==> (iii)

Tonio

3. Originally Posted by tonio
Of course this is enough, but you must know:

i) If H,K are two sbgps. of a group G, then HK is again a sbgp. iff HK = KH

ii) If N is a NORMAL sbgp. of a group G, then NK is always a sbgp. of G for any sbgp. K of G

iii) For any two sbgps. of an ABELIAN group G, HK is again a sbgp. of G.

iv) For H,K sbgps. of G, |HK| = |H||K|/|H /\ K|

Take some minutes to fully understand the above and then try to put things together.
Note also that (i ==> (ii) ==> (iii)

Tonio
Thanks Tonio. Will read what all you wrote carefully and attempt it.

4. Originally Posted by tonio
Of course this is enough, but you must know:

i) If H,K are two sbgps. of a group G, then HK is again a sbgp. iff HK = KH

ii) If N is a NORMAL sbgp. of a group G, then NK is always a sbgp. of G for any sbgp. K of G

iii) For any two sbgps. of an ABELIAN group G, HK is again a sbgp. of G.

iv) For H,K sbgps. of G, |HK| = |H||K|/|H /\ K|

Take some minutes to fully understand the above and then try to put things together.
Note also that (i ==> (ii) ==> (iii)

Tonio

Thanks a lot ! I think I got it. Let me write it and you can plz check.

I understand - "Note also that (i ==> (ii) ==> (iii)" In an abelian G every sub-group is normal.

Right, so here is my approach

1. Prime Factorization of d - Let $d = p_1^{a_1}p_2^{a_2}.....p_k^{a_k}$

2. So we have sub-gorups $H_1, H_2....H_k$ of order $p_1^{a_1},p_2^{a_2},.....,p_k^{a_k}$ respectively

3. Claim $H = H_1H_2....H_k$ is the desired sub-group of order = d
a. That it is a subgorup is obvious from your point (iii)
b. To prove $|H| = p_1^{a_1}p_2^{a_2}.....p_k^{a_k}$, I will use induction.
Basic logic of induction
Consider $H=H_1H_2$
$H_1\cap H_2$ is a sub-group of both $H_1$ and $H_2$. But $gcd(H_1,H_2)=1$, thus $|H_1\cap H_2|$ =1
Hence the result by your point# (iv)

I think I have got it. If you could plz let me know if I missed/mistaked somewhere.

And, thanks for your help.

5. Originally Posted by aman_cc
Thanks a lot ! I think I got it. Let me write it and you can plz check.

I understand - "Note also that (i ==> (ii) ==> (iii)" In an abelian G every sub-group is normal.

Right, so here is my approach

1. Prime Factorization of d - Let $d = p_1^{a_1}p_2^{a_2}.....p_k^{a_k}$

2. So we have sub-gorups $H_1, H_2....H_k$ of order $p_1^{a_1},p_2^{a_2},.....,p_k^{a_k}$ respectively

3. Claim $H = H_1H_2....H_k$ is the desired sub-group of order = d
a. That it is a subgorup is obvious from your point (iii)
b. To prove $|H| = p_1^{a_1}p_2^{a_2}.....p_k^{a_k}$, I will use induction.
Basic logic of induction
Consider $H=H_1H_2$
$H_1\cap H_2$ is a sub-group of both $H_1$ and $H_2$. But $gcd(H_1,H_2)=1$, thus $|H_1\cap H_2|$ =1
Hence the result by your point# (iv)

I think I have got it. If you could plz let me know if I missed/mistaked somewhere.

And, thanks for your help.

Great! I'd give full marks for such a response.

Tonio