# Show that the set {0} with multiplication is a group.

I'm just a little confused, I thought the identity for multiplication is always 1, yet I was looking at this problem online and it says that 0 is the identity for this problem, this was taken from "Math is fun":

"Show that the set {0} with multiplication is a group. For any elements a and b of {0}, (a*b) is an element of {0}. The closure law has been followed."

"For any a,b,c of {0}; a*(bc) = (ab)c. The associative law has been followed. For any a of {0} ia=a, where i is a particular element in {0}.The left identity element i is 0 here.

For any a of {0} the equation x*a=i has a solution known as the left inverse of a.0 is the only element in {0} and the left inverse of 0 is 0. All these properties are followed by this set that is closed under multiplication.

Therefore, {0} is a group with respect to multiplication."

This was someones answer to the problem, and I found it very confusing, and I did read up on the theory behind it.

• The only element you have is $0$. Does $0$ respects all of these properties? Jun 19, 2017 at 10:48
• So? I don't understand where the problem is. Explain better so I can help. Jun 19, 2017 at 10:50
• Sorry, basically, I thought the identity is always 1 for multiplication, but it says above in the answer I found that the identity is 0....why? Jun 19, 2017 at 10:52
• @MichaelO'Driscoll Because in this situation the only thing you can multiply with is $0$ and then always $0x=x$ (since $x=0$ always). Jun 19, 2017 at 10:57
• Since the problem seem to be that you're confused you should probably explain what confuses you. Otherwise it would be hard for us to unconfuse you. Jun 19, 2017 at 10:59

Closure:

Is it true that in the set $\{0\}$, for any two elements $a,b\in\{0\}$, the product $a\times b$ is also an element of $\{0\}$?

Yes! Proof:

• If $a\in\{0\}$, then $a=0$
• If $b\in\{0\}$, then $b=0$.
• Therefore, $a\cdot b=0\cdot 0=0$.
• Therefore, because $0\in\{0\}$, we conclude $a\cdot b\in\{0\}$.

For associativity, a very similar argument can be made.

Identity:

Is $0$ the identity of $\{0\}$? That is, is it true that for any element $a\in\{0\}$, the element $a\cdot 0=a$?

Yes! Proof:

• If $a\in\{0\}$, then $a=0$.
• Therefore, $a\cdot 0 = 0\cdot 0=0$.
• Since $a=0$ and $a\cdot 0=0$, we conclude $a\cdot 0=a$.
• Thanks, that makes sense, but what about the identity? I would assume it is 1 immediately Jun 19, 2017 at 10:55
• @MichaelO'Driscoll See my added answer. $1$ cannot be the identity because $1$ is not an element of $\{0\}$.
– 5xum
Jun 19, 2017 at 10:56
• I see, apologies...just learning this stuff by myself, thanks for your answer Jun 19, 2017 at 10:57
• @MichaelO'Driscoll Don't apologize, you asked a legitimate question. It's designed to be a little confusing, but it teaches an important point (i.e., that sometimes, $0$ can be a multiplicative identity, as long as $0$ is the only number you can multiply by).
– 5xum
Jun 19, 2017 at 10:59

1 and 0 are just names that we use to communicate. It doesn't really matter how you call your elements, it only matters what they do.