Let G be a group and |G| = 77. Let a, b ∈ G such that a ≠ b ≠ e, and |<a>| ≠ |<b>|. By Lagrange's Theorem, |<a>| divides |G| and |<b>| divides |G|. Since 7 and 11 are the only numbers that divide 77, they must be the orders of <a> and <b>. So the only subgroup containing a and b is G. Or a and b are the only elements of G, is that right?
But if either a or b generates the whole 77 elements, then one of them must be order one, right? So they're still the only elements of G.
They can't be of order 1 if they generate 77 elements: that's counter-definitorial. Likewise with them being the only elements of G, since G has 77 elements.
What I gave for an answer is that there are only two ways to factor 77, namely 7x11 and 1x77. So a and b must be order 7 and 11 (or vice-versa), or 1 and 77 (or vice-versa). Since if there is another element in G, say c, which isn't the identity and has order ≠ |a| ≠ |b|, then |c| also divides |G|, and 77 has three factors. But it doesn't, it only has two, so a and b are the only elements in G and the only subgroup of G that contains both is G.
You haven't given the question you were asked... (And it's still wrong to state that a and b are the only elements of G. Can you see why?)
Says who? What I really should say about |G| = 77 is that if a or b is not the identity, then neither can be order one. So |G|/|a| = |b|, and <a> and <b> are the only subgroups of G. Obviously G has 77 elements.
What funkstar is saying is that you haven't told us what question you were asked (the one you were referring to when you said you "answered" so and so). Why is |G|/|a| = |b|? Why can't a or b be of order 77? For example, consider the group {1, -1, i, -i} {fourth roots of unity) under multiplication. Both i and -1 are distinct non-identity elements, with |<i>| = 4 and |<-1>| = 2. So i generates the entire group. So in your example, isn't it possible that a (say) is of order 77, so that <a> = G, and b is of order 7 or 11?
You know, you have a knack for asking really basic homework questions, and then being somewhat of a prick to the people who try to help you. You should stop that.
When the thread began- it looked like a homework question to me. Most forums I've been on (And kicked off of) have a pretty strong stance against doing homework for posters.
Exactly. If G is the group generated by element-by-element multiplication of the pair \(\left( e^{\tiny \frac{2 \pi}{7} i }, \; e^{\tiny \frac{2 \pi}{11} i } \right)\) Then 1 element of G generates a group of size 1; 6 elements of G generate a group of size 7; 10 elements of G generate a group of size 11; and 60 elements of G generate a group of size 77 (i.e. G). The same is true if G is the group of generates by multiplication of \(e^{\tiny \frac{2 \pi}{77} i}\) with the sets of interesting generators: \(e^{\tiny \frac{22 \pi}{77} i }, e^{\tiny \frac{44 \pi}{77} i }, e^{\tiny \frac{66 \pi}{77} i }, e^{\tiny \frac{88 \pi}{77} i }, e^{\tiny \frac{110 \pi}{77} i }, e^{\tiny \frac{132 \pi}{77} i }\) and \(e^{\tiny \frac{14 \pi}{77} i }, e^{\tiny \frac{28 \pi}{77} i }, e^{\tiny \frac{42 \pi}{77} i }, e^{\tiny \frac{56 \pi}{77} i }, e^{\tiny \frac{110 \pi}{70} i }, e^{\tiny \frac{84 \pi}{77} i }, e^{\tiny \frac{98 \pi}{77} i }, e^{\tiny \frac{112 \pi}{77} i }, e^{\tiny \frac{126 \pi}{77} i }, e^{\tiny \frac{140 \pi}{77} i }\). Saying |<a>| ≠ |<b>| does not force |<b>| ≠ 77.
Suppose H ≤ G, and a,b ∈ H. The pair |a|,|b| must be in {7,11,77} and distinct since a ≠ b ≠ e. If |a| or |b| = 77 then |H| = |G|. If |a| = 7 and |b| = 11, then since |a| divides |H| and |b| divides |H|, |H| must be 77 and |H| = |G|. So the only subgroup of G that contains both a and b is G. Yeah, and I have this problem with being overly sensitive to criticism. . .
That's right. You didn't tell us that was what you were trying to prove. (You only gave a slightly incorrect proof of it in the original post). The above proof is correct (as it takes care of the case where |a| or |b| is 77, unlike your first proof).
