# A is a subset of B

• Oct 9th 2011, 11:03 AM
dwsmith
A is a subset of B
Prove that if A is a subset of B, then $\leq $.

Let $|A|=n \ \text{and} \ |B|=m$

$m\geq n$

$=\{e,a,a^2,\cdots, a^{n-1}\}$

$=\{e,b,b^2,\cdots, b^{m-1}\}$

Can I just since |A| less than or equal to |B|, <A> is a subgroup of <B>?
• Oct 10th 2011, 02:26 AM
Swlabr
Re: A is a subset of B
Quote:

Originally Posted by dwsmith
Prove that if A is a subset of B, then $\leq $.

Let $|A|=n \ \text{and} \ |B|=m$

$m\geq n$

$=\{e,a,a^2,\cdots, a^{n-1}\}$

$=\{e,b,b^2,\cdots, b^{m-1}\}$

Can I just since |A| less than or equal to |B|, <A> is a subgroup of <B>?

What is the written down-definition for $\langle X\rangle$? This all follows quite quickly once you know this...
• Oct 10th 2011, 07:38 AM
HallsofIvy
Re: A is a subset of B
Frankly what you have written makes little sense. If A and B are sets then " $\{e, a, a^2, ..., a^{n-1}\}$" is meaningless because there is no operation in a set. Apparently you are talking about groups, not sets. But even then you have to define <A>. You appear to mean a sub-group generated by a single element of the group but which element? Or are you assuming from the start that A and B are themselves generated by a single element?
• Oct 10th 2011, 07:47 AM
Swlabr
Re: A is a subset of B
Quote:

Originally Posted by HallsofIvy
Frankly what you have written makes little sense. If A and B are sets then " $\{e, a, a^2, ..., a^{n-1}\}$" is meaningless because there is no operation in a set. Apparently you are talking about groups, not sets. But even then you have to define <A>. You appear to mean a sub-group generated by a single element of the group but which element? Or are you assuming from the start that A and B are themselves generated by a single element?

What he means is $A\subset B\subset G$, and he requires to prove that $\langle A\rangle\leq \langle B\rangle$. However, this is clear by the very definition of $\langle X\rangle$!