A is a subset of B

**dwsmith** · October 9th 2011, 11:03 AM

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

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

$m\geq n$

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

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

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

**Swlabr** · October 10th 2011, 02:26 AM

Originally Posted by dwsmith

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

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

$m\geq n$

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

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

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

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

**HallsofIvy** · October 10th 2011, 07:38 AM

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?

**Swlabr** · October 10th 2011, 07:47 AM

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 $\langle X\rangle$ !

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