# Groups and Subgroups

• Oct 19th 2012, 07:12 AM
mous99
Groups and Subgroups
Let Q be the set of rational numbers and H = { 5m/7^n | m, n are integers and n > o}.
Determine whether H is subgroups of <Q, +>.
• Oct 19th 2012, 08:12 AM
FernandoRevilla
Re: Groups and Subgroups
Quote:

Originally Posted by mous99
Let Q be the set of rational numbers and H = { 5m/7^n | m, n are integers and n > o}. Determine whether H is subgroups of <Q, +>.

Verify (i) $H\neq \emptyset$. (ii) If $m,n,m',n'$ integers with $n>0,n'>0$, then $\dfrac{5m}{7^n}-\dfrac{5m'}{7^{n'}}=\dfrac{5k}{7^q}$ with $k,q$ integers and $q>0$.
• Oct 19th 2012, 08:24 AM
rayman
Re: Groups and Subgroups
Quote:

Originally Posted by FernandoRevilla
Verify (i) $H\neq \emptyset$. (ii) If $m,n,m',n'$ integers with $n>0,n'>0$, then $\dfrac{5m}{7^n}-\dfrac{5m'}{7^{n'}}=\dfrac{5k}{7^q}$ with $k,q$ integers and $q>0$.

this is not enough. The last condition that remains is that every element in the subset has to have an additive inverse.
• Oct 19th 2012, 08:40 AM
FernandoRevilla
Re: Groups and Subgroups
Quote:

Originally Posted by rayman
this is not enough. The last condition that remains is that every element in the subset has to have an additive inverse.

Those are necessary and sufficient conditions, according to a well known therorem (characterization of subroups).
• Oct 19th 2012, 10:53 AM
rayman
Re: Groups and Subgroups
Not according to my textbook. I am not sure what ''well known theorem'' you are referring to but according to subgroup criterion that is in my textbook (Durbin ''modern algebra'') in order to investigate whether a given subset is a subgroup we need to see it these criterions are satisfied
1) set should not be empty
2) it should be closed with respect to the operation
3) every element of the set should have an inverse with respect to the operation
• Oct 19th 2012, 11:34 AM
Bingk
Re: Groups and Subgroups
Rayman, it's commonly known as the one-step subgroup test. You'll notice that in Fernando's second statement, it has a minus, and not a plus.
In any case, it's easy enough to see that the additive inverse will be from -m.
i.e. $\frac{5m}{7^n} + \frac{5(-m)}{7^n} = 0$
• Oct 19th 2012, 11:36 AM
rayman
Re: Groups and Subgroups
and that completes the subgroup criterion :)
• Oct 19th 2012, 01:39 PM
FernandoRevilla
Re: Groups and Subgroups
Quote:

Originally Posted by rayman
Not according to my textbook. I am not sure what ''well known theorem'' you are referring to but according to subgroup criterion that is in my textbook (Durbin ''modern algebra'') in order to investigate whether a given subset is a subgroup we need to see it these criterions are satisfied
1) set should not be empty
2) it should be closed with respect to the operation
3) every element of the set should have an inverse with respect to the operation

For if you are interested, here is proof that those three conditions are equivalent to only two: PlanetMath
• Oct 20th 2012, 09:30 AM
mous99
Re: Groups and Subgroups
Quote:

Originally Posted by FernandoRevilla
Verify (i) $H\neq \emptyset$. (ii) If $m,n,m',n'$ integers with $n>0,n'>0$, then $\dfrac{5m}{7^n}-\dfrac{5m'}{7^{n'}}=\dfrac{5k}{7^q}$ with $k,q$ integers and $q>0$.

How to verify (i)?
• Oct 21st 2012, 04:32 AM
FernandoRevilla
Re: Groups and Subgroups
Quote:

Originally Posted by mous99
How to verify (i)?

Choose for example, $m=0$ and $n=1$. Then, $\dfrac{5\cdot 0}{7^1}=0\in H$.