Thread: A is a subset of B

Prove that if A is a subset of B, then $\displaystyle <A>\leq <B>$.

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

$\displaystyle m\geq n$

$\displaystyle <A>=\{e,a,a^2,\cdots, a^{n-1}\}$

$\displaystyle <B>=\{e,b,b^2,\cdots, b^{m-1}\}$

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

Originally Posted by dwsmith

Prove that if A is a subset of B, then $\displaystyle <A>\leq <B>$.

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

$\displaystyle m\geq n$

$\displaystyle <A>=\{e,a,a^2,\cdots, a^{n-1}\}$

$\displaystyle <B>=\{e,b,b^2,\cdots, b^{m-1}\}$

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

What is the written down-definition for $\displaystyle \langle X\rangle$? This all follows quite quickly once you know this...

Frankly what you have written makes little sense. If A and B are sets then "$\displaystyle \{e, a, a^2, ..., a^{n-1}\}$" 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?

Originally Posted by HallsofIvy

Frankly what you have written makes little sense. If A and B are sets then "$\displaystyle \{e, a, a^2, ..., a^{n-1}\}$" 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 $\displaystyle A\subset B\subset G$, and he requires to prove that $\displaystyle \langle A\rangle\leq \langle B\rangle$. However, this is clear by the very definition of $\displaystyle \langle X\rangle$!