Find all the subgroups of G = (Z6, *), where * is the usual multiplication sign.

[wild agate]

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[wintry juniper]

Assuming Z6 is the set of equivalence classes modulo 6 then it doesn’t form a group with multiplication ( or you are maybe referring to the group of invertibles of Z6)?

Okay well it doesn’t form a group with multiplication because for example 2 isn’t invertible are you sure it isn’t referring to (Z6)*? Basically the set of invertibles of the ring Z6

If it’s with addition though it’s indeed a cyclic group

Okay I see

Yea but if you consider (Z6,*) with * denoting multiplication then it isn’t a group that’s the problem ( because not all elements have an inverse) but if you consider the set of invertibles (Z6)x then it’s indeed a group with multiplication

And (Z6)x={1,5} so it’s just a group of cardinal 2

(5 is it’s own inverse because it’s order is 2)

Are you sure * isn’t addition ?

Well that’s not a group, I I think I’m misunderstanding

No

In Zn an element m has an inverse for multiplication if and only if Gcd(m,n)=1

( it’s basically Bezhout’s theorem)

But for example Gcd(2,6)=2 so 2 isn’t invertible

(Well with equivalence classes in mind so Bar(2))

Yes

Well Z6 is a cyclic group

So all the subgroups are cyclic as well

Well here take an element and determine it’s inverse in (Z6,+) then that element and it’s inverse and 0 form a subgroup for example {0,1,5} forms a subgroup

Here the cardinal of the group is still 6 so lagrange’s theorem still applies

So you can try searching every subgroup of cardinal 1 then 2 then 3 then 6

[mossy brook]

@jolly rock has given 1 rep to @wintry juniper

[wintry juniper]

You’re welcome !

[empty swift]

{0,1,5} is not a group, neither for + nor for *

[wintry juniper]

Here it’s in Z6 so 1+5=0 mod(6)

It’s not a subgroup of R but it is for Z6

It’s not addition in the usual sense here its class equivalences

It’s not clear here we should be noting « bar(1) » instead of 1

But it’s very annoying to type that

We can even replace 5 with -1

If we want, both are equal in terms of equivalence classes

[empty swift]

What I meant is:
1+1 = 2 ∉ {0,1,5}
0 admits no inverse in ({0,1,5},*)

@past anvil To find the other subgroups, you can look for elements which have an order that divides the cardinal of the group.

[mossy brook]

@jolly rock has given 1 rep to @empty swift

[empty swift]

No he did not mess up. The order of an element in a group is defined with the law defining the structure of group. Notice that 1 is of order 6 because 1+1+1+1+1+1=6•1=0 and for 0<k<6, k•1 ≠0

[wintry juniper]

Oh yeah you are right I’m so dumb lmao

my brain was turned off

[night socket]

{1}
{1,5}
{1,3}
{1,3,5}
These are not subgroups of (Z_6, +)

This is not a group since there are elements which do not have a multiplicative inverse, namely those which are not coprime with 6

[empty swift]

We are working with addition, not multiplication.

That's my understanding after reading the comments, though the wording says otherwise.

[mossy brook]

@jolly rock has given 1 rep to @wintry juniper @empty swift @night socket