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Here's the full problem:

Show whether the groups $G_{10},G_{7}$ are cyclic or not. If so, find their generators.

$G_{10}$ and $G_7$ are the sets of invertible elements mod 10 and mod 7. So, I found that both $G_{10},G_{7}$ were cyclic under addition with all elements contained in each set being generators and I found that only $G_7$ was cyclic under multiplication. My question then:

If it is not given which binary operation we're dealing with, is a group then cyclic if it is cyclic under both addition and multiplication?

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    $\begingroup$ Huh? $G_{10},G_7$ are not groups under addition, they have no identity and are not closed under the group operation. $\endgroup$ – Adam Hughes Aug 6 '14 at 17:51
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    $\begingroup$ They are under product. Also, a cyclic group under addition can't be cyclic under product because it won't even be a group (the "$0$" is not invertible) $\endgroup$ – Darth Geek Aug 6 '14 at 17:54
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    $\begingroup$ @DarthGeek it can if the set is $\{0\}$, see my answer below. $\endgroup$ – Adam Hughes Aug 6 '14 at 17:57
  • $\begingroup$ Yeah I realized after that neither group contained 0 and hence not groups under addition. I still found that $G_7$ was the only cyclic group. Thank you, guys! $\endgroup$ – user146925 Aug 8 '14 at 14:09
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In particular, you can see that $G_{10},G_7$ are not groups under addition, because they have no identity and are not closed under the group operation (as before). If you're always dealing which look like subsets of "stuff modulo $n$" for some $n$, then it can never be a group under both operations, unless it's $\{0\}$.

To see this, let $S$ be a subset of a set of residue classes modulo $n$ and assume $S$ is a group under addition. Clearly $0$ must be in the set, since that is the identity for such a set as an additive group, but then if $S\ne\{0\}$ we have some $0\ne x\in S$ and no matter what $x$ is, $0x=0$ so $0$ has no multiplicative inverse in the set, and so it cannot be a group for multiplication since it violates the inverses axiom.

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